Applications of Definite Integration

Find the area of the region bounded by the curve $y = 2x - x^2$ and the line $y = -x$.

Solving the equation of the curve $y = 2x - x^2$ and the equation of the line $y = - x$ $\;$ simultaneously gives the points of intersection of the two as $\left(0,0\right)$ and $\left(3,-3\right)$

The curve $y = 2x - x^2$ cuts the X axis at the point $\left(2,0\right)$ in the first quadrant.

Let the area enclosed by the curve $y = 2x - x^2$ and the X axis in the first quadrant $= A_1 = \left|I_1\right|$ where

$\begin{aligned} I_1 & = \int \limits_{0}^{2} y \; dx \\\\ & = \int \limits_{0}^{2} \left(2x - x^2\right) \; dx \\\\ & = 2 \times \left[\dfrac{x^2}{2}\right]_{0}^{2} - \left[\dfrac{x^3}{3}\right]_{0}^{2} \\\\ & = 4 - \dfrac{8}{3} = \dfrac{4}{3} \end{aligned}$

$\therefore$ $\;$ $A_1 = \left|I_1\right| = \dfrac{4}{3}$ sq units $\;\;\; \cdots \; (1)$

Let the area enclosed by the line $y = -x$, the X axis and the line $x = 3$ be $A_2 = \left|I_2\right|$ where

$\begin{aligned} I_2 & = \int \limits_{0}^{3} y \; dx \\\\ & = \int \limits_{0}^{3} -x \; dx \\\\ & = \left[\dfrac{- x^2}{2}\right]_{0}^{3} = \dfrac{-9}{2} \end{aligned}$

$\therefore$ $\;$ $A_2 = \left|I_2\right| = \dfrac{9}{2}$ sq units $\;\;\; \cdots \; (2)$

Let the area enclosed by the curve $y = 2x - x^2$, the X axis and the line $x = 3$ be $A_3 = \left|I_3\right|$ where

$\begin{aligned} I_3 & = \int \limits_{2}^{3} y \; dx \\\\ & = \int \limits_{2}^{3} \left(2x - x^2\right) \; dx \\\\ & = 2 \times \left[\dfrac{x^2}{2}\right]_{2}^{3} - \left[\dfrac{x^3}{3}\right]_{2}^{3} \\\\ & = 9 - 4 - \dfrac{1}{3} \left(27 - 8\right) = \dfrac{-4}{3} \end{aligned}$

$\therefore$ $\;$ $A_3 = \left|I_3\right| = \dfrac{4}{3}$ sq units $\;\;\; \cdots \; (3)$

$\therefore$ $\;$ Area of the region bounded by the curve $y = 2x - x^2$ and the line $y = -x$ $\;$ (shaded region) is

$\begin{aligned} A & = A_1 + A_2 - A_3 \\\\ & = \dfrac{4}{3} + \dfrac{9}{2} - \dfrac{4}{3} \hspace{3em} \left[\text{From equations } (1), (2) \text{ and } (3)\right] \\\\ & = \dfrac{9}{2} \; \text{sq units} \end{aligned}$

Applications of Definite Integration

Find the area enclosed by the circle $x^2 + y^2 = a^2$

$x^2 + y^2 = a^2$ $\;\;\; \cdots \; (1)$

is a circle with center at $\left(0,0\right)$ and radius $= a$ units.

The circle cuts the positive X and Y axis at the points $\left(a,0\right)$ and $\left(0,a\right)$ respectively.

Required area $= A = 4 \left|I\right|$ where

$I = \displaystyle \int \limits_{0}^{a} y \; dx$ $\;\;\; \cdots \; (2)$

We have from equation $(1)$, $y = \sqrt{a^2 - x^2}$ $\;\;\; \cdots \; (3)$

$\therefore$ $\;$ In view of equation $(3)$, equation $(2)$ becomes

$I = \displaystyle \int \limits_{0}^{a} \sqrt{a^2 - x^2} \; dx$ $\;\;\; \cdots \; (4)$

Let $x = a \sin \theta$ $\;\;\; \cdots \; (5)$

Differentiating equation $(5)$ gives

$dx = a \cos \theta \; d\theta$ $\;\;\; \cdots \; (5a)$

Also, we have from equation $(5)$, $\theta = \sin^{-1} \left(\dfrac{x}{a}\right)$

$\text{When} \begin{cases} x = 0, & \theta = \sin^{-1} \left(0\right) = 0 \\ x = a, & \theta = \sin^{-1} \left(1\right) = \dfrac{\pi}{4} \end{cases}$ $\;\;\; \cdots \; (5b)$

$\therefore$ $\;$ In view of equations $(5)$, $(5a)$ and $(5b)$, equation $(4)$ can be written as

$\begin{aligned} I & = \int \limits_{0}^{\frac{\pi}{2}} \sqrt{a^2 - a^2 \sin^2 \theta} \times a \; \cos \theta \; d \theta \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} a \sqrt{1 - \sin^2 \theta} \times a \; \cos \theta \; d \theta \\\\ & = a^2 \int \limits_{0}^{\frac{\pi}{2}} \cos \theta \times \cos \theta \; d \theta \\\\ & = a^2 \int \limits_{0}^{\frac{\pi}{2}} \cos^2 \theta \; d \theta \\\\ & = a^2 \int \limits_{0}^{\frac{\pi}{2}} \left(\dfrac{1 + \cos 2 \theta}{2}\right) \; d \theta \\\\ & = \dfrac{a^2}{2} \left\{\int \limits_{0}^{\frac{\pi}{2}} d \; \theta + \int \limits_{0}^{\frac{\pi}{2}} \cos 2 \theta \; d \theta \right\} \\\\ & = \dfrac{a^2}{2} \left\{\left[\theta\right]_{0}^{\pi / 2} + \dfrac{1}{2} \left[\sin 2 \theta\right]_{0}^{\pi / 2} \right\} \\\\ & = \dfrac{a^2}{2} \left\{\dfrac{\pi}{2} - 0 + \dfrac{1}{2} \left[\sin \left(\pi\right) - \sin \left(0\right)\right] \right\} \\\\ & = \dfrac{\pi \; a^2}{4} + 0 = \dfrac{\pi \; a^2}{4} \end{aligned}$

$\therefore$ $\;$ Required area $= 4 \times \dfrac{\pi \; a^2}{4} = \pi a^2$ square units

Applications of Definite Integration

Find the area bounded by the X axis and the parabola $y = -x^2 + 4$

Given: $y = -x^2 + 4$ $\;\;\; \cdots \; (1)$

$\implies$ $x^2 = - \left(y - 4\right)$ $\;\;\; \cdots \; (2)$

Equation $(2)$ represents a parabola.

It cuts the X axis at the points $\left(\pm 2 , 0\right)$.

Its vertex is $\left(0,4\right)$.

Equation of X axis is $y = 0$.

$\therefore$ $\;$ Required area (shaded portion) $= A = 2 \left|I\right|$ where

$I = \displaystyle \int \limits_{0}^{4} x \; dy$ $\;\;\; \cdots \; (3)$

From equation $(1)$, $x = \sqrt{4 - y}$ $\;\;\; \cdots \; (4)$

$\therefore$ $\;$ In view of equation $(4)$, equation $(3)$ can be written as

$\begin{aligned} I & = \int \limits_{0}^{4} \sqrt{4 - y} \; dy \\\\ & = \dfrac{-2}{3} \left[\left(4 - y\right)^{3/2}\right]_{0}^{4} \\\\ & = \dfrac{-2}{3} \left[0 - \left(4\right)^{3/2}\right] \\\\ & = \dfrac{2}{3} \times 8 = \dfrac{16}{3} \end{aligned}$

$\therefore$ $\;$ Required area $= 2 I = \dfrac{32}{3}$ square units

Applications of Definite Integration

Find the area bounded by the parabola $y = x^2 - 4$, the X axis and the lines $x = -1$ and $x = 2$.

Given: $y = x^2 - 4$ $\;\;\; \cdots \; (1)$

Required area (shaded portion in the figure) $= \left|I\right|$ where

$\begin{aligned} I & = \int \limits_{-1}^{2} y \; dx \\\\ & = \int \limits_{-1}^{2} \left(x^2 - 4\right) \; dx \hspace{3em} \left[\text{From equation }(1) \right] \\\\ & = \left[\dfrac{x^3}{3} - 4 x\right]_{-1}^{2} \\\\ & = \dfrac{2^3}{3} - 4 \times 2 - \dfrac{\left(-1\right)^3}{3} + 4 \times \left(-1\right) \\\\ & = - 9 \end{aligned}$

$\therefore$ Required area $= \left|I\right| = 9$ square units

Definite Integration

Prove that $\displaystyle \int \limits_{0}^{\frac{\pi}{2}} \left[2 \log \left(\sin x\right) - \log \left(\sin 2x\right)\right] \; dx = - \dfrac{\pi}{2} \log \left(2\right)$

Let $I = \displaystyle \int \limits_{0}^{\frac{\pi}{2}} \left[2 \log \left(\sin x\right) - \log \left(\sin 2x\right)\right] \; dx$ $\;\;\; \cdots \; (1)$

Now,

$\begin{aligned} 2 \log \left(\sin x\right) - \log \left(\sin 2x\right) & = \log \left(\sin^2 x\right) - \log \left(2 \sin x \cos x\right) \\\\ & = \log \left(\dfrac{\sin^2 x}{2 \sin x \cos x}\right) \\\\ & \hspace{3em} \left[\text{Note: } \log \left(C\right) - \log \left(D\right) = \log \left(\dfrac{C}{D}\right)\right] \\\\ & = \log \left(\dfrac{\sin x}{2 \cos x}\right) \\\\ & = \log \left(\dfrac{\tan x}{2}\right) \\\\ & = \log \left(\tan x\right) - \log \left(2\right) \;\;\; \cdots \; (2) \end{aligned}$

$\therefore$ $\;$ In view of equation $(2)$, equation $(1)$ can be written as

$I = \displaystyle \int \limits_{0}^{\frac{\pi}{2}} \log \left(\tan x\right) \; dx - \int \limits_{0}^{\frac{\pi}{2}} \log \left(2\right) \; dx$ $\;\;\; \cdots \; (3)$

$\begin{aligned} \text{Let } I_1 & = \int \limits_{0}^{\frac{\pi}{2}} \log \left(\tan x\right) \; dx \;\;\; \cdots \; (4) \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \log \left[\tan \left(\dfrac{\pi}{2} - x\right)\right] \; dx \hspace{3em} \left[\text{Note: } \int \limits_{0}^{a} f\left(x\right) \; dx = \int \limits_{0}^{a} f\left(a - x\right) \; dx\right] \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \log \left(\cot x\right) \; dx \;\;\; \cdots \; (5) \end{aligned}$

Adding equations $(4)$ and $(5)$ gives

$\begin{aligned} 2 I_1 & = \int \limits_{0}^{\frac{\pi}{2}} \left[\log \left(\tan x\right) + \log \left(\cot x\right)\right] \; dx \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \log \left(\tan x \times \cot x\right) \; dx \hspace{3em} \left[\text{Note: } \log C + \log D = \log \left(C \cdot D\right)\right] \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \log \left(1\right) \; dx = 0 \end{aligned}$

$\implies$ $I_1 = 0$ $\;\;\; \cdots \; (6)$

$\begin{aligned} \text{Let } I_2 & = \int \limits_{0}^{\frac{\pi}{2}} \log \left(2\right) \; dx \\\\ & = \log \left(2\right) \times \left[x\right]_{0}^{\pi / 2} \end{aligned}$

i.e. $I_2 = \dfrac{\pi}{2} \log \left(2\right)$ $\;\;\; \cdots \; (7)$

$\therefore$ $\;$ In view of equations $(6)$ and $(7)$, equation $(3)$ becomes

$I = 0 - \dfrac{\pi}{2} \log \left(2\right) = - \dfrac{\pi}{2} \log \left(2\right)$ $\hspace{2em}$ Hence proved

Definite Integration

Evaluate $\displaystyle \int \limits_{0}^{\frac{\pi}{2}} \left|\sin x - \cos x\right| \; dx$

Let $I = \displaystyle \int \limits_{0}^{\frac{\pi}{2}} \left|\sin x - \cos x\right| \; dx$ $\;\;\; \cdots \; (1)$

$\text{Now, } \left|\sin x - \cos x\right| = \begin{cases} \cos x - \sin x, & 0 < x < \dfrac{\pi}{4} \\\\ \sin x - \cos x, & \dfrac{\pi}{4} < x < \dfrac{\pi}{2} \end{cases}$ $\;\;\; \cdots \; (2)$

$\therefore$ $\;$ In view of equation $(2)$, equation $(1)$ can be written as

$\begin{aligned} I & = \int \limits_{0}^{\frac{\pi}{4}} \left(\cos x - \sin x\right) \; dx + \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \left(\sin x - \cos x\right) \; dx \\\\ & = \int \limits_{0}^{\frac{\pi}{4}} \cos x \; dx - \int \limits_{0}^{\frac{\pi}{4}} \sin x \; dx + \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin x \; dx - \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos x \; dx \\\\ & = \left[\sin x\right]_{0}^{\pi / 4} + \left[\cos x\right]_{0}^{\pi / 4} - \left[\cos x\right]_{\pi / 4}^{\pi / 2} - \left[\sin x\right]_{\pi / 4}^{\pi / 2} \\\\ & = \sin \left(\dfrac{\pi}{4}\right) - \sin \left(0\right) + \cos \left(\dfrac{\pi}{4}\right) - \cos \left(0\right) \\\\ & \hspace{5em} - \cos \left(\dfrac{\pi}{2}\right) + \cos \left(\dfrac{\pi}{4}\right) - \sin \left(\dfrac{\pi}{2}\right) + \sin \left(\dfrac{\pi}{4}\right) \\\\ & = \dfrac{1}{\sqrt{2}} - 0 + \dfrac{1}{\sqrt{2}} - 1 - 0 + \dfrac{1}{\sqrt{2}} - 1 + \dfrac{1}{\sqrt{2}} \\\\ & = \dfrac{4}{\sqrt{2}} - 2 \\\\ & = 2 \sqrt{2} - 2 = 2 \left(\sqrt{2} - 1\right) \end{aligned}$

Definite Integration

Evaluate $\displaystyle \int \limits_{0}^{\pi} \dfrac{x^2 \; \sin x \; dx}{\left(2x - \pi\right) \left(1 + \cos^2 x\right)}$

$\begin{aligned} \text{Let } I & = \int \limits_{0}^{\pi} \dfrac{x^2 \; \sin x \; dx}{\left(2x - \pi\right) \left(1 + \cos^2 x\right)} \;\;\; \cdots \; (1) \\\\ & = \int \limits_{0}^{\pi} \dfrac{\left(\pi - x\right)^2 \; \sin \left(\pi - x\right) \; dx}{\left\{2 \left(\pi - x\right) - \pi \right\} \left\{1 + \left[\cos \left(\pi - x\right)\right]^2 \right\}} \\\\ & \hspace{2em} \left[\text{Note: } \int \limits_{0}^{a} f\left(x\right) \; dx = \int \limits_{0}^{a} f\left(a - x\right) \; dx\right] \\\\ & = \int \limits_{0}^{\pi} \dfrac{\left(\pi^2 - 2 \pi x + x^2\right) \; \sin x \; dx}{\left(\pi - 2x\right) \left(1 + \cos^2 x\right)} \;\;\; \cdots \; (2) \end{aligned}$

$\therefore$ $\;$ Adding equations $(1)$ and $(2)$ gives

$\begin{aligned} 2 I & = \int \limits_{0}^{\pi} \left[\dfrac{x^2 \; \sin x}{\left(2x - \pi\right) \left(1 + \cos^2 x\right)} + \dfrac{\left(\pi^2 - 2 \pi x + x^2\right) \; \sin x}{\left(\pi - 2x\right) \left(1 + \cos^2 x\right)}\right] \; dx \\\\ & = \int \limits_{0}^{\pi} \left[\dfrac{x^2 \; \sin x - \left(\pi^2 - 2 \pi x + x^2\right) \; \sin x}{\left(2x - \pi\right) \left(1 + \cos^2 x\right)}\right] \; dx \\\\ & = \int \limits_{0}^{\pi} \dfrac{\left(2 \pi x - \pi^2\right) \sin x}{\left(2x - \pi\right) \left(1 + \cos^2 x\right)} \; dx \\\\ & = \int \limits_{0}^{\pi} \dfrac{\pi \left(2x - \pi\right) \sin x}{\left(2x - \pi\right) \left(1 + \cos^2 x\right)} \; dx \\\\ & = \pi \int \limits_{0}^{\pi} \dfrac{\sin x \; dx}{1 + \cos^2 x} \;\;\; \cdots \; (3) \end{aligned}$

Let $\cos x = u$ $\;\;\; \cdots \; (4)$

Differentiating equation $(4)$ gives

$- \sin x \; dx = du$ $\implies$ $\sin x \; dx = - du$ $\;\;\; \cdots \; (4a)$

$\text{When } \begin{cases} x = 0, & u = \cos \left(0\right) = 1 \\ x = \pi, & u = \cos \left(\pi\right) = -1 \end{cases}$ $\;\;\; \cdots \; (4b)$

$\therefore$ $\;$ In view of equations $(4)$, $(4a)$ and $(4b)$, equation $(3)$ can be written as

$2 I = \pi \displaystyle \int \limits_{1}^{-1} \dfrac{-du}{1 + u^2}$

$\begin{aligned} \text{i.e. } I & = \dfrac{\pi}{2} \int \limits_{-1}^{1} \dfrac{du}{1 + u^2} \hspace{3em} \left[\text{Note: } \int \limits_{b}^{a} -dx = \int \limits_{a}^{b} dx\right] \\\\ & = \dfrac{\pi}{2} \left[\tan^{-1} \left(u\right)\right]_{-1}^{1} \\\\ & = \dfrac{\pi}{2} \left[\tan^{-1} \left(1\right) - \tan^{-1} \left(-1\right)\right] \\\\ & = \dfrac{\pi}{2} \left[\dfrac{\pi}{4} - \left(\dfrac{- \pi}{4}\right)\right] \\\\ & = \dfrac{\pi}{2} \times \dfrac{\pi}{2} = \dfrac{\pi^2}{4} \end{aligned}$

Definite Integration

Evaluate $\displaystyle \int \limits_{0}^{1} \dfrac{\log \left(1 + t\right)}{1 + t^2} \; dt$

Let $I = \displaystyle \int \limits_{0}^{1} \dfrac{\log \left(1 + t\right)}{1 + t^2} \; dt$ $\;\;\; \cdots \; (1)$

Put $\tan^{-1}\left(t\right) = x$ $\;\;\; \cdots \; (2)$

$\implies$ $t = \tan \left(x\right)$ $\;\;\; \cdots \; (2a)$

Differentiating equation $(2)$ gives

$\dfrac{dt}{1 + t^2} = dx$ $\;\;\; \cdots \; (2b)$

$\text{When } \begin{cases} t = 0, & x = \tan^{-1} \left(0\right) = 0 \\ t = 1, & x = \tan^{-1} \left(1\right) = \dfrac{\pi}{4} \end{cases}$ $\;\;\; \cdots \; (2c)$

$\therefore$ $\;$ In view of equations $(2)$, $(2a)$, $(2b)$ and $(2c)$, equation $(1)$ can be written as

$\begin{aligned} I & = \int \limits_{0}^{\frac{\pi}{4}} \log \left[1 + \tan x\right] \; dx \\\\ & = \int \limits_{0}^{\frac{\pi}{4}} \log \left[1 + \tan \left(\dfrac{\pi}{4} - x\right)\right] \; dx \;\;\; \cdots \; (3) \hspace{1em} \left[\text{Note: } \int \limits_{0}^{a} f\left(x\right) \; dx = \int \limits_{0}^{a} f\left(a - x\right) \; dx\right] \end{aligned}$

$\begin{aligned} \text{Now, } \tan\left(\dfrac{\pi}{4} - x\right) & = \dfrac{\tan \left(\dfrac{\pi}{4}\right) - \tan \left(x\right)}{1 + \tan \left(\dfrac{\pi}{4}\right) \times \tan \left(x\right)} \\\\ & = \dfrac{1 - \tan \left(x\right)}{1 + \tan \left(x\right)} \\\\ \therefore \; 1 + \tan \left(\dfrac{\pi}{4} - x\right) & = 1 + \dfrac{1 - \tan \left(x\right)}{1 + \tan \left(x\right)} \\\\ & = \dfrac{1 + \tan \left(x\right) + 1 - \tan \left(x\right)}{1 + \tan \left(x\right)} \\\\ & = \dfrac{2}{1 + \tan x} \;\;\; \cdots \; (4) \end{aligned}$

$\therefore$ $\;$ In view of equation $(4)$ equation $(3)$ becomes

$I = \displaystyle \int \limits_{0}^{\frac{\pi}{4}} \log \left[\dfrac{2}{1 + \tan x}\right] \; dx$

i.e. $I = \displaystyle \int \limits_{0}^{\frac{\pi}{4}} \log \left(2\right) \; dx - \int \limits_{0}^{\frac{\pi}{4}} \log \left(1 + \tan x\right) \; dx$

i.e. $I = \log \left(2\right) \displaystyle \int \limits_{0}^{\frac{\pi}{4}} dx - I$

i.e. $2 I = \log \left(2\right) \times \left[x\right]_{0}^{\pi / 4}$

i.e. $2 I = \dfrac{\pi}{4} \log \left(2\right)$

i.e. $I = \dfrac{\pi}{8} \log \left(2\right)$

Definite Integration

Evaluate $\displaystyle \int \limits_{0}^{\frac{\pi}{4}} \dfrac{\sin^2 x}{1 + \sin x \cos x} \; dx$

Let $I = \displaystyle \int \limits_{0}^{\frac{\pi}{4}} \dfrac{\sin^2 x}{1 + \sin x \cos x} \; dx$ $\;\;\; \cdots \; (1)$

Now, $\tan x = \dfrac{\sin x}{\cos x}$ $\implies$ $\sin x = \tan x \cos x$ $\implies$ $\sin x = \dfrac{\tan x}{\sec x}$ $\;\;\; \cdots \; (2a)$

and $\cos x = \dfrac{1}{\sec x}$ $\;\;\; \cdots \; (2b)$

$\therefore$ $\;$ In view of equations $(2a)$ and $(2b)$, equation $(1)$ becomes

$\begin{aligned} I & = \int \limits_{0}^{\frac{\pi}{4}} \dfrac{\tan^2 x / sec^2 x}{1 + \dfrac{\tan x}{\sec x} \times \dfrac{1}{\sec x}} \; dx \\\\ & = \int \limits_{0}^{\frac{\pi}{4}} \dfrac{\tan^2 x}{\sec^2 x + \tan x} \; dx \\\\ & = \int \limits_{0}^{\frac{\pi}{4}} \dfrac{\tan^2 x}{1 + \tan^2 x + \tan x} \; dx \;\;\; \cdots \; (3) \end{aligned}$

Let $\tan x = u$ $\;\;\; \cdots \; (4)$

Differentiating equation $(4)$ gives

$\sec^2 x \; dx = du$

$\implies$ $dx = \dfrac{du}{\sec^2 x} = \dfrac{du}{1 + \tan^2 x} = \dfrac{du}{1 + u^2}$ $\;\;\; \cdots \; (4a)$

$\text{When } \begin{cases} x = 0, & u = \tan \left(0\right) = 0 \\ x = \dfrac{\pi}{4}, & u = \tan \left(\dfrac{\pi}{4}\right) = 1 \end{cases}$ $\;\;\; \cdots \; (4b)$

$\therefore$ $\;$ In view of equations $(4)$, $(4a)$ and $(4b)$, equation $(3)$ becomes

$I = \displaystyle \int \limits_{0}^{1} \dfrac{u^2 \; du}{\left(1 + u + u^2\right) \left(1 + u^2\right)}$ $\;\;\; \cdots \; (5)$

Let $\dfrac{u^2}{\left(1 + u^2\right) \left(1 + u + u^2\right)} = \dfrac{Au + B}{1 + u^2} + \dfrac{Cu + D}{1 + u + u^2}$ $\;\;\; \cdots \; (6)$

i.e. $u^2 = \left(A u + B\right) \left(1 + u + u^2\right) + \left(C u + D\right) \left(1 + u^2\right)$

i.e. $u^2 = Au + Au^2 + Au^3 + B + Bu + Bu^2 + Cu + Cu^3 + D + Du^2$

i.e. $u^2 = u^3 \left(A + C\right) + u^2 \left(A + B + D\right) + u \left(A + B + D\right) + \left(B + D\right)$

Comparing the coefficient of $u^3$ term gives

$A + C = 0$ $\;\;\; \cdots \; (6a)$

Comparing the coefficient of $u^2$ term gives

$A + B + D = 1$ $\;\;\; \cdots \; (6b)$

Comparing the coefficient of $u$ term gives

$A + B + C = 0$ $\;\;\; \cdots \; (6c)$

Comparing the constant term gives

$B + D = 0$ $\;\;\; \cdots \; (6d)$

We have from equations $(6b)$ and $(6d)$, $A = 1$

Substituting the value of A in equation $(6a)$ gives $C = - A = -1$

Substituting the values of A and C in equation $(6c)$ gives $1 + B - 1 = 0$ $\implies$ $B = 0$

Substituting the value of B in equation $(6d)$ gives $D = 0$

$\therefore$ $\;$ Substituting the values of A, B, C and D, equation $(6)$ we have,

$\dfrac{u^2}{\left(1 + u^2\right) \left(1 + u + u^2\right)} = \dfrac{u}{1 + u^2} - \dfrac{u}{1 + u + u^2}$ $\;\;\; \cdots \; (7)$

$\therefore$ $\;$ In view of equation $(7)$ equation $(5)$ can be written as

$\begin{aligned} I & = \int \limits_{0}^{1} \dfrac{u \; du}{u^2 + 1} - \int \limits_{0}^{1} \dfrac{u \; du}{u^2 + u + 1} \\\\ & = I_1 - I_2 \;\;\; \cdots \; (8) \end{aligned}$

Consider $I_1 = \displaystyle \int \limits_{0}^{1} \dfrac{u \; du}{u^2 + 1}$ $\;\;\; \cdots \; (9)$

Let $u^2 + 1 = t$ $\;\;\; \cdots \; (10)$

Differentiating equation $(10)$ gives

$2 u \; du = dt$ $\implies$ $u \; du = \dfrac{dt}{2}$ $\;\;\; \cdots \; (10a)$

$\text{When } \begin{cases} u = 0, & t = 1 \\ u = 1, & t = 2 \end{cases}$ $\;\;\; \cdots \; (10b)$

$\therefore$ $\;$ In view of equations $(10)$, $(10a)$ and $(10b)$, equation $(9)$ becomes

$\begin{aligned} I_1 & = \dfrac{1}{2} \int \limits_{1}^{2} \dfrac{dt}{t} \\\\ & = \dfrac{1}{2} \left[\log \left|t\right|\right]_{1}^{2} \\\\ & = \dfrac{1}{2} \left[\log \left(2\right) - \log \left(1\right)\right] \\\\ & = \dfrac{1}{2} \log \left(2\right) \;\;\; \cdots \; (11) \end{aligned}$

$\begin{aligned} \text{Consider } I_2 & = \int \limits_{0}^{1} \dfrac{u \; du}{u^2 + u + 1} \\\\ & = \dfrac{1}{2} \int \limits_{0}^{1} \dfrac{2 u \; du}{u^2 + u + 1} \\\\ & = \dfrac{1}{2} \left[\int \limits_{0}^{1} \dfrac{2u + 1}{u^2 + u + 1} \; du - \int \limits_{0}^{1} \dfrac{du}{u^2 + u + 1}\right] \;\;\; \cdots \; (12) \end{aligned}$

Consider $I_3 = \displaystyle \int \limits_{0}^{1} \dfrac{2u + 1}{u^2 + u + 1} \; du$ $\;\;\; \cdots \; (13)$

Let $u^2 + u + 1 = m$ $\;\;\; \cdots \; (14)$

Differentiating equation $(14)$ gives

$\left(2u + 1\right) \; du = dm$ $\;\;\; \cdots \; (14a)$

$\text{When } \begin{cases} u = 0, & m = 1 \\ u = 1, & m = 3 \end{cases}$ $\;\;\; \cdots \; (14b)$

$\therefore$ $\;$ In view of equations $(14)$, $(14a)$ and $(14b)$, equation $(13)$ becomes

$\begin{aligned} I_3 & = \int \limits_{1}^{3} \dfrac{dm}{m} \\\\ & = \left[\log \left|m\right|\right]_{1}^{3} \\\\ & = \log \left(3\right) - \log \left(1\right) \\\\ & = \log \left(3\right) \;\;\; \cdots \; (15) \end{aligned}$

$\begin{aligned} \text{Consider } I_4 & = \int \limits_{0}^{1} \dfrac{du}{u^2 + u + 1} \\\\ & = \int \limits_{0}^{1} \dfrac{du}{\left(u^2 + u + \dfrac{1}{4}\right) + 1 - \dfrac{1}{4}} \\\\ & = \int \limits_{0}^{1} \dfrac{du}{\left(u + \dfrac{1}{2}\right)^2 + \left(\dfrac{\sqrt{3}}{2}\right)^2} \\\\ & = \dfrac{2}{\sqrt{3}} \left[\tan^{-1} \left(\dfrac{u + \dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}\right)\right]_{0}^{1} \\\\ & \hspace{3em} \left[\text{Note: } \int \limits_{p}^{q} \dfrac{dx}{x^2 + a^2} = \dfrac{1}{a} \left[\tan^{-1} \left(\dfrac{x}{a}\right)\right]_{p}^{q}\right] \\\\ & = \dfrac{2}{\sqrt{3}} \left[\tan^{-1} \left(\dfrac{2u + 1}{\sqrt{3}}\right)\right]_{0}^{1} \\\\ & = \dfrac{2}{\sqrt{3}} \left[\tan^{-1} \left(\dfrac{3}{\sqrt{3}}\right) - \tan^{-1} \left(\dfrac{1}{\sqrt{3}}\right)\right]_{0}^{1} \\\\ & = \dfrac{2}{\sqrt{3}} \left[\dfrac{\pi}{3} - \dfrac{\pi}{6}\right] \\\\ & = \dfrac{2}{\sqrt{3}} \times \dfrac{\pi}{6} = \dfrac{\pi}{3 \sqrt{3}} \;\;\; \cdots \; (16) \end{aligned}$

$\therefore$ $\;$ In view of equations $(15)$ and $(16)$, equation $(12)$ becomes

$I_2 = \dfrac{1}{2} \left[\log \left(3\right) - \dfrac{\pi}{3 \sqrt{3}}\right] = \dfrac{1}{2} \log \left(3\right) - \dfrac{\pi}{6 \sqrt{3}}$ $\;\;\; \cdots \; (17)$

$\therefore$ $\;$ In view of equations $(11)$ and $(17)$, equation $(8)$ becomes

$\begin{aligned} I & = \dfrac{1}{2} \log \left(2\right) - \dfrac{1}{2} \log \left(3\right) + \dfrac{\pi}{6 \sqrt{3}} \\\\ & = \dfrac{\pi}{6 \sqrt{3}} - \dfrac{1}{2} \log \left(\dfrac{3}{2}\right) \end{aligned}$

Definite Integration

Evaluate $\displaystyle \int \limits_{0}^{\frac{\pi}{4}} \dfrac{dx}{a^2 \cos^2 x - b^2 \sin^2 x}$ $\hspace{3em}$ $\left(a > b > 0\right)$

$\begin{aligned} \text{Let } I & = \int \limits_{0}^{\frac{\pi}{4}} \dfrac{dx}{a^2 \cos^2 x - b^2 \sin^2 x} \\\\ & = \int \limits_{0}^{\frac{\pi}{4}} \dfrac{\dfrac{dx}{\cos^2 x}}{a^2 - \dfrac{b^2 \sin^2 x}{\cos^2 x}} \\\\ & = \int \limits_{0}^{\frac{\pi}{4}} \dfrac{\sec^2 x \; dx}{a^2 - b^2 \tan^2 x} \\\\ & = \dfrac{1}{b^2} \int \limits_{0}^{\frac{\pi}{4}} \dfrac{\sec^2 x \; dx}{\dfrac{a^2}{b^2} - \tan^2 x} \;\;\; \cdots \; (1) \end{aligned}$

Let $\tan x = u$ $\;\;\; \cdots \; (2)$

Differentiating equation $(2)$ gives

$\sec^2 x \; dx = du$ $\;\;\; \cdots \; (2a)$

$\text{When } \begin{cases} x = 0, & u = \tan \left(0\right) = 0 \\ x = \dfrac{\pi}{4}, & u = \tan \left(\dfrac{\pi}{4}\right) = 1 \end{cases}$ $\;\;\; \cdots \; (2b)$

$\therefore$ $\;$ In view of equations $(2)$, $(2a)$ and $(2b)$, equation $(1)$ becomes

$\begin{aligned} I & = \dfrac{1}{b^2} \int \limits_{0}^{1} \dfrac{du}{\dfrac{a^2}{b^2} - u^2} \\\\ & = \dfrac{1}{b^2} \int \limits_{0}^{1} \dfrac{du}{\left(a / b\right)^2 - \left(u\right)^2} \\\\ & = \dfrac{1}{b^2} \times \dfrac{b}{2a} \left[\log \left|\dfrac{\dfrac{a}{b} + u}{\dfrac{a}{b} - u}\right|\right]_{0}^{1} \hspace{3em} \left[\text{Note: }\int \limits_{p}^{q} \dfrac{dx}{a^2 - x^2} = \dfrac{1}{2a} \left[\log\left|\dfrac{a + x}{a - x}\right|\right]_{p}^{q}\right] \\\\ & = \dfrac{1}{2ab} \left[\log \left|\dfrac{\dfrac{a}{b} + 1}{\dfrac{a}{b} - 1}\right| - \log \left|\dfrac{a / b}{a / b}\right|\right] \\\\ & = \dfrac{1}{2ab} \log \left|\dfrac{a + b}{a - b}\right| \end{aligned}$

Definite Integration

Evaluate $\displaystyle \int \limits_{0}^{\frac{\pi}{2}} \log \left(\cos x\right) \; dx$

$\begin{aligned} \text{Let } I & = \int \limits_{0}^{\frac{\pi}{2}} \log \left(\cos x\right) \; dx \;\;\; \cdots \; (1) \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \log \left[\cos \left(\dfrac{\pi}{2} - x\right)\right] \; dx \hspace{3em} \left[\text{Note: } \int \limits_{0}^{a} f\left(x\right) \; dx = \int \limits_{0}^{a} f\left(a - x\right) \; dx\right] \\\\ \text{i.e. } I & = \int \limits_{0}^{\frac{\pi}{2}} \log \left(\sin x\right) \; dx \;\;\; \cdots \; (2) \end{aligned}$ Adding equations $(1)$ and $(2)$ gives

$\begin{aligned} 2 I & = \int \limits_{0}^{\frac{\pi}{2}} \left[\log \left(\cos x\right) + \log \left(\sin x\right)\right] \; dx \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \log \left(\cos x \cdot \sin x\right) \; dx \hspace{3em} \left[\text{Note: } \log C + \log D = \log \left(C \cdot D\right)\right]\\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \log \left(\dfrac{2 \sin x \cos x}{2}\right) \; dx \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \log \left(\dfrac{\sin 2x}{2}\right) \; dx \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \log \left(\sin 2x\right) \; dx - \int \limits_{0}^{\frac{\pi}{2}} \log \left(2\right) \; dx \hspace{3em} \left[\text{Note: } \log C - \log D = \log \left(\dfrac{C}{D}\right)\right] \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \log \left(\sin 2x\right) \; dx - \log \left(2\right) \times \left[x\right]_{0}^{\pi / 2} \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \log \left(\sin 2x\right) \; dx - \dfrac{\pi}{2} \; \log \left(2\right) \;\;\; \cdots \; (3) \end{aligned}$ Let $I_1 = \displaystyle \int \limits_{0}^{\frac{\pi}{2}} \log \left(\sin 2x\right) \; dx$ $\;\;\; \cdots \; (4)$

Let $2x = u$ $\;\;\; \cdots \; (5)$

Differentiating equation $(5)$ gives

$2 \; dx = du$ $\implies$ $dx = \dfrac{du}{2}$ $\;\;\; \cdots \; (5a)$

$\text{When } \begin{cases} x = 0, & u = 0 \\ x = \dfrac{\pi}{2}, & u = \pi \end{cases}$ $\;\;\; \cdots \; (5b)$

$\therefore$ $\;$ In view of equations $(5)$, $(5a)$ and $(5b)$, equation $(4)$ can be written as

$I_1 = \dfrac{1}{2} \displaystyle \int \limits_{0}^{\pi} \log \left(\sin u\right) \; du$ $\;\;\; \cdots \; (6)$

Since $\log \left[\sin \left(\pi - u\right)\right] = \log \left(\sin u\right)$

$\implies$ $\displaystyle \int \limits_{0}^{\pi} \log \left(\sin u\right) \; du = 2 \int \limits_{0}^{\frac{\pi}{2}} \log \left(\sin u\right) \; du$ $\;\;\; \cdots \; (7)$

$\therefore$ $\;$ In view of equation $(7)$ equation $(6)$ becomes

$\begin{aligned} I_1 & = \dfrac{1}{2} \times 2 \int \limits_{0}^{\frac{\pi}{2}} \log \left(\sin u\right) \; du \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \log \left(sin u\right) \; du \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \log \left(sin x\right) \; dx \hspace{3em} \left[\text{Note: Definite integral is independent of variable}\right] \\\\ & = I \hspace{3em} \left[\text{By equation }(2)\right] \;\;\; \cdots \; (8) \end{aligned}$

$\therefore$ $\;$ In view of equation $(8)$, equation $(3)$ becomes

$2 I = I - \dfrac{\pi}{2} \log \left(2\right)$

i.e. $I = - \dfrac{\pi}{2} \log \left(2\right)$

Definite Integration

Prove that $\displaystyle \int \limits_{0}^{\pi} x \; f\left(\sin x\right) \; dx = \dfrac{\pi}{2} \int \limits_{0}^{\pi} f \left(\sin x\right) \; dx$. Hence evaluate $\displaystyle \int \limits_{0}^{\pi} \dfrac{x}{1 + \sin x} \; dx$

$\begin{aligned} \text{Let } I & = \displaystyle \int \limits_{0}^{\pi} x \; f\left(\sin x\right) \; dx \;\;\; \cdots \; (1) \\\\ & = \int \limits_{0}^{\pi} \left(\pi - x\right) f \left[\sin \left(\pi - x\right)\right] \; dx \hspace{3em} \left[\text{Note: } \int \limits_{0}^{a} f\left(x\right) \; dx = \int \limits_{0}^{a} f\left(a - x\right) \; dx\right] \\\\ & = \int \limits_{0}^{\pi} \left(\pi - x\right) f \left(\sin x\right) \; dx \\\\ & = \pi \int \limits_{0}^{\pi} f \left(\sin x\right) \; dx - \int \limits_{0}^{\pi} x \; f \left(\sin x\right) \; dx \\\\ & = \pi \int \limits_{0}^{\pi} f \left(\sin x\right) \; dx - I \end{aligned}$

i.e. $2 I = \pi \displaystyle \int \limits_{0}^{\pi} f \left(\sin x\right) \; dx$

$\therefore$ $\;$ $I = \dfrac{\pi}{2} \displaystyle \int \limits_{0}^{\pi} f \left(\sin x\right) \; dx$ $\;\;\; \cdots \; (2)$ $\hspace{3em}$ Hence proved

$\begin{aligned} \text{Now, } \int \limits_{0}^{\pi} \dfrac{x}{1 + \sin x} \; dx & = \dfrac{\pi}{2} \int \limits_{0}^{\pi} \dfrac{1}{1 + \sin x} \; dx \;\;\; \left[\text{by equation } (2)\right] \\\\ & = \dfrac{\pi}{2} \int \limits_{0}^{\pi} \dfrac{1 - \sin x}{1 - \sin^2 x} \; dx \\\\ & = \dfrac{\pi}{2} \int \limits_{0}^{\pi} \dfrac{1 - \sin x}{\cos^2 x} \; dx \\\\ & = \dfrac{\pi}{2} \int \limits_{0}^{\pi} \sec^2 x \; dx - \dfrac{\pi}{2} \int \limits_{0}^{\pi} \sec x \; \tan x \; dx \\\\ & = \dfrac{\pi}{2} \left\{\left[\tan x\right]_{0}^{\pi} - \left[\sec x\right]_{0}^{\pi} \right\} \\\\ & = \dfrac{\pi}{2} \left\{\left[\tan \pi - \tan 0\right] - \left[\sec \pi - \sec 0\right] \right\} \\\\ & = \dfrac{\pi}{2} \left\{0 - \left(- 1 - 1\right) \right\} = \pi \end{aligned}$

Definite Integration

Evaluate $\displaystyle \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\sin^2 x}{\sin x + \cos x} \; dx$

$\begin{aligned} \text{Let } I & = \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\sin^2 x}{\sin x + \cos x} \; dx \;\;\; \cdots \; (1) \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\left[\sin \left(\dfrac{\pi}{2} - x\right)\right]^2}{\sin \left(\dfrac{\pi}{2} - x\right) + \cos \left(\dfrac{\pi}{2} - x\right)} \; dx \\\\ & \hspace{3em} \left[\text{Note: } \int \limits_{0}^{a} f\left(x\right) \; dx = \int \limits_{0}^{a} f\left(x - a\right) \; dx\right] \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\cos^2 x}{\cos x + \sin x} \; dx \;\;\; \cdots \; (2) \end{aligned}$

Adding equations $(1)$ and $(2)$ gives

$2 I = \displaystyle \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\sin^2 x + \cos^2 x}{\sin x + \cos x} \; dx$

i.e. $I = \dfrac{1}{2} \displaystyle \int \limits_{0}^{\frac{\pi}{2}} \dfrac{1}{\sin x + \cos x} \; dx$ $\;\;\; \cdots \; (3)$

Let $\tan \left(\dfrac{x}{2}\right) = u$ $\;\;\; \cdots \; (4)$

Differentiating equation $(4)$ gives

$\dfrac{1}{2} \sec^2 \left(\dfrac{x}{2}\right) \; dx = du$

i.e. $dx = \dfrac{2 \; du}{\sec^2 \left(\dfrac{x}{2}\right)} = \dfrac{2 \; du}{1 + \tan^2 \left(\dfrac{x}{2}\right)} = \dfrac{2 \; du}{1 + u^2}$ $\;\;\; \cdots \; (4a)$

Now, $\sin x = \dfrac{2 \tan \left(\dfrac{x}{2}\right)}{1 + \tan^2 \left(\dfrac{x}{2}\right)} = \dfrac{2 u}{1 + u^2}$ $\;\;\; \cdots \; (4b)$

and $\cos x = \dfrac{1 - \tan^2 \left(\dfrac{x}{2}\right)}{1 + \tan^2 \left(\dfrac{x}{2}\right)} = \dfrac{1 - u^2}{1 + u^2}$ $\;\;\; \cdots \; (4c)$

$\text{When } \begin{cases} x = 0, & u = \tan 0 = 0 \\ x = \dfrac{\pi}{2}, & u = \tan \left(\dfrac{\pi}{4}\right) = 1 \end{cases}$ $\;\;\; \cdots \; (4d)$

$\therefore$ $\;$ In view of equations $(4)$, $(4a)$, $(4b)$, $(4c)$ and $(4d)$, equation $(3)$ can be written as

$\begin{aligned} I & = \dfrac{1}{2} \int \limits_{0}^{1} \dfrac{\dfrac{2 \; du}{1 + u^2}}{\dfrac{2 u}{1 + u^2} + \dfrac{1 - u^2}{1 + u^2}} \\\\ & = \int \limits_{0}^{1} \dfrac{du}{1 + 2u - u^2} \\\\ & = - \int \limits_{0}^{1} \dfrac{du}{\left(u^2 - 2u + 1\right) -1 -1} \\\\ & = - \int \limits_{0}^{1} \dfrac{du}{\left(u - 1\right)^2 - 2} \\\\ & = \int \limits_{0}^{1} \dfrac{du}{\left(\sqrt{2}\right)^2 - \left(u - 1\right)^2} \\\\ & = \dfrac{1}{2 \sqrt{2}} \left[\log \left|\dfrac{\sqrt{2} + u - 1}{\sqrt{2} - u + 1}\right|\right]_{0}^{1} \hspace{3em} \left[\text{Note: } \int \dfrac{dx}{a^2 - x^2} = \dfrac{1}{2a} \log \left|\dfrac{a + x}{a - x}\right| + c\right] \\\\ & = \dfrac{1}{2 \sqrt{2}} \left[\log \left|\dfrac{\sqrt{2} + 1 - 1}{\sqrt{2} - 1 + 1}\right| - \log \left|\dfrac{\sqrt{2} + 0 - 1}{\sqrt{2} - 0 + 1}\right|\right] \\\\ & = \dfrac{1}{2 \sqrt{2}} \left[\log 1 - \log \left|\dfrac{\sqrt{2} - 1}{\sqrt{2} + 1}\right|\right] \\\\ & = \dfrac{1}{2 \sqrt{2}} \left[0 + \log \left|\dfrac{\sqrt{2} + 1}{\sqrt{2} - 1}\right|\right] \\\\ & = \dfrac{1}{2 \sqrt{2}} \log \left|\dfrac{\left(\sqrt{2} + 1\right)^2}{2 - 1}\right| \\\\ & = \dfrac{1}{2 \sqrt{2}} \log \left|\left(\sqrt{2} + 1\right)^2\right| \\\\ & = \dfrac{1}{2 \sqrt{2}} \times 2 \log \left|\sqrt{2} + 1\right| \\\\ & = \dfrac{1}{\sqrt{2}} \log \left|\sqrt{2} + 1\right| \end{aligned}$

Definite Integration

Evaluate $\displaystyle \int \limits_{0}^{\pi} \dfrac{x \sin x \; dx}{1 + \cos^2 x}$

$\begin{aligned} \text{Let } I & = \displaystyle \int \limits_{0}^{\pi} \dfrac{x \sin x \; dx}{1 + \cos^2 x} \\\\ & = \displaystyle \int \limits_{0}^{\pi} \dfrac{\left(\pi - x\right) \sin \left(\pi - x\right) \; dx}{1 + \cos^2 \left(\pi - x\right)} \hspace{3em} \left[\text{Note: } \int \limits_{0}^{a} f\left(x\right) \; dx = \int \limits_{0}^{a} f\left(x - a\right) \; dx\right] \\\\ & = \displaystyle \int \limits_{0}^{\pi} \dfrac{\left(\pi - x\right) \sin x \; dx}{1 + \left(- \cos x\right)^2} \\\\ & = \pi \displaystyle \int \limits_{0}^{\pi} \dfrac{\sin x \; dx}{1 + \cos^2 x} - \displaystyle \int \limits_{0}^{\pi} \dfrac{x \sin x \; dx}{1 + \cos^2 x} \\\\ & = \pi \displaystyle \int \limits_{0}^{\pi} \dfrac{\sin x \; dx}{1 + \cos^2 x} - I \\\\ \text{i.e. } 2 I & = \pi \displaystyle \int \limits_{0}^{\pi} \dfrac{\sin x \; dx}{1 + \cos^2 x} \\\\ & = \pi I_1 \;\;\; \cdots \; (1) \end{aligned}$

Consider $I_1 = \displaystyle \int \limits_{0}^{\pi} \dfrac{\sin x \; dx}{1 + \cos^2 x}$ $\;\;\; \cdots \; (2)$

Let $\cos x = u$ $\;\;\; \cdots \; (3)$

Differentiating equation $(3)$ gives

$- \sin x \; dx = du$ $\;\;\; \cdots \; (3a)$

$\text{When } \begin{cases} x = 0, & u = \cos 0 = 1 \\ x = \pi, & u = \cos \pi = - 1 \end{cases}$ $\;\;\; \cdots \; (3b)$

$\therefore$ $\;$ In view of equations $(3)$, $(3a)$ and $(3b)$, equation $(2)$ becomes

$\begin{aligned} I_1 & = \int \limits_{1}^{-1} \dfrac{- du}{1 + u^2} \\\\ & = \int \limits_{-1}^{1} \dfrac{du}{1 + u^2} \hspace{3em} \left[\text{Note: } - \int \limits_{b}^{a} f \left(x\right) \; dx = + \int \limits_{a}^{b} f \left(x\right) \; dx\right] \\\\ & = \left[\tan^{-1} u\right]_{-1}^{1} \\\\ & = \tan^{-1} \left(1\right) - \tan^{-1} \left(- 1\right) \\\\ & = \dfrac{\pi}{4} - \left(\dfrac{- \pi}{4}\right) = \dfrac{\pi}{2} \;\;\; \cdots \; (4) \end{aligned}$

Substituting equation $(4)$ in equation $(1)$ gives

$2 I = \pi \times \dfrac{\pi}{2}$

$\implies$ $I = \dfrac{\pi^2}{4}$

Definite Integration

Evaluate $\displaystyle \int \limits_{0}^{\pi} \dfrac{x}{1 + x} \; dx$

$\begin{aligned} \text{Let } I & = \displaystyle \int \limits_{0}^{\pi} \dfrac{x}{1 + x} \; dx \;\;\; \cdots \; (1) \\\\ & = \displaystyle \int \limits_{0}^{\pi} \dfrac{\pi - x}{1 + \sin \left(\pi - x\right)} \; dx \hspace{3em} \left[\text{Note: } \int \limits_{0}^{a} f\left(x\right) \; dx = \int \limits_{0}^{a} f\left(x - a\right) \; dx\right] \\\\ & = \int \limits_{0}^{\pi} \dfrac{\pi - x}{1 + \sin x} \; dx \;\;\; \cdots \; (2) \end{aligned}$

Adding equations $(1)$ and $(2)$ gives

$\begin{aligned} 2 I & = \int \limits_{0}^{\pi} \dfrac{x + \pi - x}{1 + \sin x} \; dx \\\\ & = \pi \int \limits_{0}^{\pi} \dfrac{dx}{1 + \sin x} \\\\ & = \pi \int \limits_{0}^{\pi} \dfrac{1 - \sin x}{1 - \sin^2 x} \; dx \\\\ & = \pi \int \limits_{0}^{\pi} \dfrac{1 - \sin x}{\cos^2 x} \; dx \end{aligned}$

$\begin{aligned} \implies I & = \dfrac{\pi}{2} \left\{\int \limits_{0}^{\pi} \dfrac{1}{\cos^2 x} \; dx - \int \limits_{0}^{\pi} \dfrac{\sin x}{\cos^2 x} \; dx \right\} \\\\ & = \dfrac{\pi}{2} \left\{\int \limits_{0}^{\pi} \sec^2 x \; dx - \int \limits_{0}^{\pi} \sec x \; \tan x \; dx \right\} \\\\ & = \dfrac{\pi}{2} \left\{\left[\tan x\right]_{0}^{\pi} - \left[\sec x\right]_{0}^{\pi} \right\} \\\\ & = \dfrac{\pi}{2} \left[\tan \pi - \tan 0 - \sec \pi + \sec 0\right] \\\\ & = \dfrac{\pi}{2} \left[0 - 0 + 1 + 1\right] \\\\ & = \dfrac{\pi}{2} \times 2 = \pi \end{aligned}$

Definite Integration

Evaluate $\displaystyle \int \limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{1}{1 + \sqrt{\cot x}} \; dx$

$\begin{aligned} \text{Let } I & = \displaystyle \int \limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{1}{1 + \sqrt{\cot x}} \; dx \;\;\; \cdots \; (1) \\\\ & = \displaystyle \int \limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{1}{1 + \sqrt{\cot \left(\dfrac{\pi}{6} + \dfrac{\pi}{3} - x\right)}} \; dx \\\\ & \left[\text{Note: } \int \limits_{a}^{b} f\left(x\right) \; dx = \int \limits_{a}^{b} f \left(a + b - x\right) \; dx\right] \\\\ & = \int \limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{1}{1 + \sqrt{\cot \left(\dfrac{\pi}{2} - x\right)}} \; dx \\\\ & = \int \limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{1}{1 + \sqrt{\tan x}} \; dx \\\\ & = \int \limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{1}{1 + \dfrac{1}{\sqrt{\cot x}}} \; dx \\\\ & = \int \limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{\sqrt{\cot x}}{\sqrt{\cot x} + 1} \; dx \;\;\; \cdots \; (2) \end{aligned}$

Adding equations $(1)$ and $(2)$ gives

$2 I = \displaystyle \int \limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{1 + \sqrt{\cot x}}{1 + \sqrt{\cot x}} \; dx$

i.e. $2 I = \int \limits_{\frac{\pi}{6}}^{\frac{\pi}{3}} dx = \left[x\right]_{\pi / 6}^{\pi / 3} = \dfrac{\pi}{3} - \dfrac{\pi}{6} = \dfrac{\pi}{6}$

$\implies$ $I = \dfrac{\pi}{12}$

Definite Integration

Evaluate $\displaystyle \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}} \; dx$

$\begin{aligned} \text{Let } I & = \displaystyle \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}} \; dx \;\;\; \cdots \; (1) \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\sqrt{\sin \left(\dfrac{\pi}{2} - x\right)}}{\sqrt{\cos \left(\dfrac{\pi}{2} - x\right)}+ \sqrt{\sin \left(\dfrac{\pi}{2} - x\right)}} \; dx \\\\ & \left[\text{Note: } \int \limits_{0}^{a} f\left(x\right) \; dx = \int \limits_{0}^{a} f\left(x - a\right) \; dx\right] \\\\ \text{i.e. } I & = \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \; dx \;\;\; \cdots \; (2) \end{aligned}$

Adding equations $(1)$ and $(2)$ gives

$\begin{aligned} 2 I & = \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \; dx \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} dx \\\\ & = \left[x\right]_{0}^{\pi / 2} = \dfrac{\pi}{2} \end{aligned}$

$\implies$ $I = \dfrac{\pi}{4}$

Definite Integration

If $\displaystyle \int \limits_{0}^{a} \sqrt{x} \; dx = 2 a \int \limits_{0}^{\frac{\pi}{2}} \sin^3 x \; dx$, then obtain $\displaystyle \int \limits_{a}^{a + 1} x \; dx$

$\displaystyle \int \limits_{0}^{a} \; \sqrt{x} \; dx = \dfrac{2}{3} \left[x^{3/2}\right]_{0}^{a} = \dfrac{2}{3} a^{3/2}$ $\;\;\; \cdots \; (1)$

$\begin{aligned} \int \limits_{0}^{\frac{\pi}{2}} \sin^3 x \; dx & = \int \limits_{0}^{\frac{\pi}{2}} \left(\dfrac{3 \sin x - \sin 3x}{4}\right) \; dx \hspace{3em} \left[\text{Note: }\sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta \right] \\\\ & = \dfrac{3}{4} \int \limits_{0}^{\frac{\pi}{2}} \sin x \; dx - \dfrac{1}{4} \int \limits_{0}^{\frac{\pi}{2}} \sin 3x \; dx \\\\ & = \dfrac{3}{4} \left[- \cos x\right]_{0}^{\pi / 2} - \dfrac{1}{4} \left[\dfrac{- \cos 3x}{3}\right]_{0}^{\pi / 2} \\\\ & = - \dfrac{3}{4} \left(0 - 1\right) + \dfrac{1}{12} \left(0 - 1\right) \\\\ & = \dfrac{3}{4} - \dfrac{1}{12} = \dfrac{2}{3} \;\;\; \cdots \; (2) \end{aligned}$

$\therefore$ $\;$ As per the question, we have from equations $(1)$ and $(2)$,

$\dfrac{2}{3} a^{3/2} = 2 a \times \dfrac{2}{3} = \dfrac{4a}{3}$

i.e. $a^{3/2 - 1} = 2$

i.e. $a^{1/2} = 2$ $\implies$ $a = 4$ $\;\;\; \cdots \; (3)$

$\therefore$ $\;$ In view of equation $(3)$ we have

$\displaystyle \int \limits_{a}^{a+1} x \; dx = \int \limits_{4}^{5} x \; dx = \left[\dfrac{x^2}{2}\right]_{4}^{5} = \dfrac{1}{2} \left(25 - 16\right) = \dfrac{9}{2}$

Definite Integration

If $\displaystyle \int \limits_{\sqrt{2}}^{k} \dfrac{dx}{x \sqrt{x^2 - 1}} \; dx = \dfrac{\pi}{12}$, obtain k.

Let $I = \displaystyle \int \limits_{\sqrt{2}}^{k} \dfrac{dx}{x \sqrt{x^2 - 1}} \; dx$ $\;\;\; \cdots \; (1)$

Put $x = \sec \theta$ $\;\;\; \cdots \; (2)$

Differentiating equation $(2)$ gives

$dx = \sec \theta \tan \theta \; d\theta$ $\;\;\; \cdots \; (2a)$

From equation $(2)$, $\theta = \sec^{-1}x$

$\therefore \; \text{When } \begin{cases} x = \sqrt{2}, & \theta = \sec^{-1} \left(\sqrt{2}\right) = \dfrac{\pi}{4} \\ x = k, & \theta = \sec^{-1} \left(k\right) \end{cases}$ $\;\;\; \cdots \; (2b)$

$\therefore$ $\;$ In view of equations $(2)$, $(2a)$ and $(2b)$, equation $(1)$ can be written as

$\begin{aligned} I & = \int \limits_{\frac{\pi}{4}}^{\sec^{-1} \left(k\right)} \dfrac{\sec \theta \; \tan \theta \; d \theta}{\sec \theta \; \sqrt{\sec^2 \theta - 1}} \\\\ & = \int \limits_{\frac{\pi}{4}}^{\sec^{-1} \left(k\right)} d \theta \\\\ & = \left[\theta\right]_{\pi / 4}^{\sec^{-1} \left(k\right)} \\\\ & = \sec^{-1} \left(k\right) - \dfrac{\pi}{4} \;\;\; \cdots \; (3) \end{aligned}$

Given $I = \dfrac{\pi}{12}$ $\;\;\; \cdots \; (4)$

$\therefore$ $\;$ We have from equations $(3)$ and $(4)$,

$\sec^{-1} \left(k\right) - \dfrac{\pi}{4} = \dfrac{\pi}{12}$

i.e. $\sec^{-1} \left(k\right) = \dfrac{\pi}{12} + \dfrac{\pi}{4} = \dfrac{\pi}{3}$

$\implies$ $k = \sec \left(\dfrac{\pi}{3}\right) = 2$

Definite Integration

Evaluate $\displaystyle \int \limits_{0}^{2 \pi} \left|\cos x\right| \; dx$

Let $I = \displaystyle \int \limits_{0}^{2 \pi} \left|\cos x\right| \; dx$ $\;\;\; \cdots \; (1)$

$\left|\cos x\right| = \begin{cases} + \cos x, & 0 < x < \dfrac{\pi}{2} \\ - \cos x, & \dfrac{\pi}{2} < x < \dfrac{3 \pi}{2} \\ + \cos x, & \dfrac{3 \pi}{2} < x < 2 \pi \end{cases}$ $\;\;\; \cdots \; (2)$

$\therefore$ $\;$ In view of equation $(2)$, equation $(1)$ can be written as

$\begin{aligned} I & = \int \limits_{0}^{\frac{\pi}{2}} \cos x \; dx - \int \limits_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} \cos x \; dx + \int \limits_{\frac{3 \pi}{2}}^{2 \pi} \cos x \; dx \\\\ & = \left[\sin x\right]_{0}^{\pi / 2} - \left[\sin x\right]_{\pi / 2}^{3 \pi / 2} + \left[\sin x\right]_{3 \pi / 2}^{2 \pi} \\\\ & = \left[\sin \left(\dfrac{\pi}{2}\right) - \sin 0\right] - \left[\sin \left(\dfrac{3 \pi}{2}\right) - \sin \left(\dfrac{\pi}{2}\right)\right] + \left[\sin \left(2 \pi\right) - \sin \left(\dfrac{3 \pi}{2}\right)\right] \\\\ & = \left(1 - 0\right) - \left(- 1 - 1\right) + \left(0 + 1\right) \\\\ & = 4 \end{aligned}$

Definite Integration

Evaluate $\displaystyle \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos \left(2x\right) \cdot \log \left(\sin x\right) \; dx$

$\begin{aligned} \text{Let } I & = \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos \left(2x\right) \cdot \log \left(\sin 2x\right) \; dx \\\\ & \left[\begin{aligned} \text{Note: }& \int \limits_{a}^{b} u\; v \; dx = \left[u \int v \; dx\right]_{a}^{b} - \int \limits_{a}^{b} \left\{\int v \; dx \times \dfrac{d}{dx} \left(du\right) \right\} \; dx \\\\ & \text{Here } u = \log \left(\sin x\right); \; v = \cos \left(2x\right) \end{aligned}\right] \\\\ & = \left[\log \left(\sin x\right) \int \cos \left(2x\right) \; dx\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} - \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \left\{\int \cos \left(2x\right) \times \dfrac{d}{dx} \left[\log \left(\sin x\right)\right] \right\} \; dx \\\\ & = \left[\dfrac{1}{2} \sin \left(2x\right) \; \log \left(\sin x\right)\right]_{\pi / 4}^{\pi / 2} - \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \dfrac{\sin \left(2x\right)}{2} \times \dfrac{1}{\sin x} \times \cos x \; dx \\\\ & = \dfrac{1}{2} \left[\sin \left(\pi\right) \; \log \left|\sin \left(\dfrac{\pi}{2}\right)\right| - \sin \left(\dfrac{\pi}{2}\right) \; \log \left|\sin \left(\dfrac{\pi}{4}\right)\right|\right] \\ & \hspace{10em} - \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \dfrac{2 \sin x \cos x \times \cos x}{2 \sin x} \; dx \\\\ & = \dfrac{1}{2} \left[0 - 1 \times \log \left|\dfrac{1}{\sqrt{2}}\right|\right] - \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos^2 x \; dx \\\\ & = \dfrac{1}{2} \log \left|\sqrt{2}\right| - \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \left(\dfrac{1 + \cos 2x}{2}\right) \; dx \\\\ & = \dfrac{1}{4} \log 2 - \dfrac{1}{2} \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} dx - \dfrac{1}{2} \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos 2x \; dx \\\\ & = \dfrac{1}{4} \log 2 - \dfrac{1}{2} \left[x\right]_{\pi / 4}^{\pi / 2} - \dfrac{1}{2} \left[\dfrac{\sin 2x}{2}\right]_{\pi / 4}^{\pi / 2} \\\\ & = \dfrac{1}{4} \log 2 - \dfrac{1}{2} \left[\dfrac{\pi}{2} - \dfrac{\pi}{4}\right] - \dfrac{1}{4} \left[\sin \left(\pi\right) - \sin \left(\dfrac{\pi}{2}\right)\right] \\\\ & = \dfrac{1}{4} \log 2 - \dfrac{\pi}{8} - \dfrac{1}{4} \left[0 - 1\right] \\\\ & = \dfrac{1}{4} \log 2 - \dfrac{\pi}{8} + \dfrac{1}{4} \end{aligned}$

Definite Integration

Evaluate $\displaystyle \int \limits_{0}^{1} \sqrt{\dfrac{1 - x}{1 + x}} \; dx$

Let $I = \displaystyle \int \limits_{0}^{1} \sqrt{\dfrac{1 - x}{1 + x}} \; dx$ $\;\;\; \cdots \; (1)$

Let $1 + x = u$ $\;\;\; \cdots \; (2)$

Differentiating equation $(2)$ gives

$dx = du$ $\;\;\; \cdots \; (2a)$

$\text{When } \begin{cases} x = 0, & u = 1 + 0 = 1 \\ x = 1, & u = 1 + 1 = 2 \end{cases}$ $\;\;\; \cdots \; (2b)$

Also, from equation $(2)$, $x = u - 1$

$\therefore$ $\;$ $1 - x = 2 - u$ $\;\;\; \cdots \; (2c)$

$\therefore$ $\;$ In view of equations $(2)$, $(2a)$, $(2b)$ and $(2c)$, equation $(1)$ can be written as

$I = \displaystyle \int \limits_{1}^{2} \dfrac{\sqrt{2 - u}}{\sqrt{u}} \; du$ $\;\;\; \cdots \; (3)$

Let $\sqrt{u} = v$ $\;\;\; \cdots \; (4)$

Differentiating equation $(4)$ gives

$\dfrac{1}{2 \sqrt{u}} \; du = dv$

$\implies$ $\dfrac{du}{\sqrt{u}} = 2 \; dv$ $\;\;\; \cdots \; (4a)$

$\text{When } \begin{cases} u = 1, & v = \sqrt{1} = 1 \\ u = 2, & v = \sqrt{2} \end{cases}$ $\;\;\; \cdots \; (4b)$

Also, from equation $(4)$, $u = v^2$ $\;\;\; \cdots \; (4c)$

$\therefore$ $\;$ In view of equations $(4)$, $(4a)$, $(4b)$ and $(4c)$, equation $(3)$ can be written as

$I = 2 \displaystyle \int \limits_{1}^{\sqrt{2}} \sqrt{2 - v^2} \; dv$ $\;\;\; \cdots \; (5)$

Let $v = \sqrt{2} \sin \theta$ $\;\;\; \cdots \; (6)$

Differentiating equation $(6)$ gives

$dv = \sqrt{2} \cos \theta \; d \theta$ $\;\;\; \cdots \; (6a)$

$\text{When } \begin{cases} v = 1, & \theta = \sin^{-1} \left(\dfrac{1}{\sqrt{2}}\right) = \dfrac{\pi}{4} \\ v = \sqrt{2}, & \theta = \sin^{-1} \left(\dfrac{\sqrt{2}}{\sqrt{2}}\right) = \dfrac{\pi}{2} \end{cases}$ $\;\;\; \cdots \; (6b)$

Also, from equation $(6)$, $\theta = \sin^{-1} \left(\dfrac{v}{\sqrt{2}}\right)$ $\;\;\; \cdots \; (6c)$

$\therefore$ $\;$ In view of equations $(6)$, $(6a)$, $(6b)$ and $(6c)$, equation $(5)$ can be written as

$\begin{aligned} I & = 2 \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sqrt{2 - 2 \sin^2 \theta} \times \sqrt{2} \cos \theta \; d \theta \\\\ & = 2 \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sqrt{2} \left(\sqrt{1 - \sin^2 \theta}\right) \times \sqrt{2} \cos \theta \; d \theta \\\\ & = 4 \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos \theta \times \cos \theta \; d \theta \\\\ & = 4 \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos^2 \theta \; d \theta \\\\ & = 4 \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \left(\dfrac{1 + \cos 2 \theta}{2}\right) \; d \theta \\\\ & = 2 \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} d \theta + 2 \int \limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos 2 \theta \; d \theta \\\\ & = 2 \left[\theta\right]_{\pi / 4}^{\pi / 2} + 2 \left[\dfrac{\sin 2 \theta}{2}\right]_{\pi / 4}^{\pi / 2} \\\\ & = 2 \left[\dfrac{\pi}{2} - \dfrac{\pi}{4}\right] + \left[\sin \left(\pi\right) - \sin \left(\dfrac{\pi}{2}\right)\right] \\\\ & = \dfrac{\pi}{2} + \left(0 - 1\right) \\\\ & = \dfrac{\pi}{2} - 1 \end{aligned}$

Definite Integration

Evaluate $\displaystyle \int \limits_{0}^{1} \sin^{-1} \left(\sqrt{\dfrac{x}{x + 1}}\right) \; dx$

$\begin{aligned} \text{Let } I & = \int \limits_{0}^{1} \sin^{-1} \left(\sqrt{\dfrac{x}{x + 1}}\right) \; dx \\\\ & \left[\begin{aligned} \text{Note: }& \int \limits_{a}^{b} u\; v \; dx = \left[u \int v \; dx\right]_{a}^{b} - \int \limits_{a}^{b} \left\{\int v \; dx \times \dfrac{d}{dx} \left(du\right) \right\} \; dx \\\\ & \text{Here } u = \sin^{-1} \left(\sqrt{\dfrac{x}{x + 1}}\right); \; v = 1 \end{aligned}\right] \\\\ & = \left[\sin^{-1} \left(\sqrt{\dfrac{x}{x + 1}}\right) \int dx\right]_{0}^{1} \\ & \hspace{5em} - \int \limits_{0}^{1} \left\{\int dx \times \dfrac{d}{dx} \left[\sin^{-1} \left(\sqrt{\dfrac{x}{x + 1}}\right)\right] \right\} \; dx \;\;\; \cdots \; (1) \end{aligned}$

$\begin{aligned} \text{Now, } \dfrac{d}{dx} \left[\sin^{-1} \left(\sqrt{\dfrac{x}{x + 1}}\right)\right] & = \dfrac{1}{\sqrt{1 - \dfrac{x}{x + 1}}} \times \dfrac{d}{dx} \left(\sqrt{\dfrac{x}{x + 1}}\right) \\\\ & = \sqrt{x + 1} \times \left(\dfrac{\sqrt{x + 1} \times \dfrac{1}{2 \sqrt{x}} - \sqrt{x} \times \dfrac{1}{2 \sqrt{x + 1}}}{x + 1}\right) \\\\ & = \sqrt{x + 1} \times \left(\dfrac{x + 1 - x}{2 \left(x + 1\right) \sqrt{x} \sqrt{x + 1}}\right) \\\\ & = \dfrac{1}{2 \left(x + 1\right) \sqrt{x}} \;\;\; \cdots \; (2) \end{aligned}$

$\therefore$ $\;$ In view of equation $(2)$, equation $(1)$ becomes

$\begin{aligned} I & = \left[x \; \sin^{-1} \left(\sqrt{\dfrac{x}{x + 1}}\right)\right]_{0}^{1} - \dfrac{1}{2} \int \limits_{0}^{1} \dfrac{x \; dx}{\left(x + 1\right) \sqrt{x}} \\\\ & = 1 \times \sin^{-1} \left(\sqrt{\dfrac{1}{2}}\right) - 0 - \dfrac{1}{2} \int \limits_{0}^{1} \dfrac{x \; dx}{\left(x + 1\right) \sqrt{x}} \;\;\; \cdots \; (3) \end{aligned}$

Consider $\displaystyle \int \limits_{0}^{1} \dfrac{\sqrt{x}}{x + 1} \; dx$ $\;\;\; \cdots \; (4)$

Put $\sqrt{x} = u$ $\;\;\; \cdots \; (5)$

Differentiating equation $(5)$ gives

$\dfrac{1}{2 \sqrt{x}} \; dx = du$ $\implies$ $dx = 2 \sqrt{x} \; du = 2 u \; du$ $\;\;\; \cdots \; (5a)$

$\text{When } \begin{cases} x = 0, & u = \sqrt{0} = 0 \\ x = 1, & u = \sqrt{1} = 1 \end{cases}$ $\;\;\; \cdots \; (5b)$

$\therefore$ $\;$ In view of equations $(5)$, $(5a)$ and $(5b)$, equation $(4)$ becomes

$\begin{aligned} \int \limits_{0}{1} \dfrac{\sqrt{x}}{x + 1} \; dx & = \int \limits_{0}^{1} \dfrac{u \times 2u}{u^2 + 1} \; du \\\\ & = 2 \int \limits_{0}^{1} \dfrac{u^2}{u^2 + 1} \; du \\\\ & = 2 \left[\int \limits_{0}^{1} \dfrac{u^2 + 1}{u^2 + 1} \; du - \int \limits_{0}^{1} \dfrac{du}{u^2 + 1}\right] \\\\ & = 2 \left[\int \limits_{0}^{1} du - \int \limits_{0}^{1} \dfrac{du}{u^2 + 1}\right] \\\\ & = 2 \left\{\left[u\right]_{0}^{1} - \left[\tan^{-1} u\right]_{0}^{1} \right\} \\\\ & = 2 \left[1 - 0 - \tan^{-1} \left(1\right) + \tan^{-1} \left(0\right)\right] \\\\ & = 2 \left[1 - \dfrac{\pi}{4}\right] \\\\ & = 2 - \dfrac{\pi}{2} \;\;\; \cdots \; (6) \end{aligned}$

$\therefore$ $\;$ Substituting equation $(6)$ in equation $(3)$ gives

$\begin{aligned} I & = \dfrac{\pi}{4} - \dfrac{1}{2} \left(2 - \dfrac{\pi}{2}\right) \\\\ & = \dfrac{\pi}{4} - 1 + \dfrac{\pi}{4} = \dfrac{\pi}{2} - 1 \end{aligned}$

Definite Integration

Evaluate $\displaystyle \int \limits_{0}^{\frac{\pi}{2}} x^2 \; \cos 2x \; dx$

$\begin{aligned} \text{Let } I & = \int \limits_{0}^{\frac{\pi}{2}} x^2 \; \cos 2x \; dx \\\\ & \left[\begin{aligned} \text{Note: }& \int \limits_{a}^{b} u\; v \; dx = \left[u \int v \; dx\right]_{a}^{b} - \int \limits_{a}^{b} \left\{\int v \; dx \times \dfrac{d}{dx} \left(du\right) \right\} \; dx \\\\ & \text{Here } u = x^2; \; v = \cos 2x \end{aligned}\right] \\\\ & = \left[x^2 \int \cos 2x \; dx\right]_{0}^{\frac{\pi}{2}} - \int \limits_{0}^{\frac{\pi}{2}} \left[\int \cos 2x \; dx \times \dfrac{d}{dx} \left(x^2\right)\right] dx \\\\ & = \dfrac{1}{2} \left[x^2 \; \sin 2x\right]_{0}^{\pi / 2} - \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\sin 2x}{2} \times 2 x \; dx \\\\ & = \dfrac{1}{2} \left[\dfrac{\pi^2}{4} \; \sin \left(\pi\right) - 0\right] - \int \limits_{0}^{\frac{\pi}{2}} x \; \sin \left(2x\right) \; dx \\\\ & \hspace{10em} \left[\text{Here } u = x; \; v = \sin \left(2x\right)\right] \\\\ & = 0 - \left[x \int \sin \left(2x\right) \; dx\right]_{0}^{\pi / 2} + \int \limits_{0}^{\frac{\pi}{2}} \left[\sin \left(2x\right) \; dx \times \dfrac{d}{dx} \left(x\right)\right] \; dx \\\\ & = \dfrac{1}{2} \left[x \; \cos \left(2x\right)\right]_{0}^{\pi / 2} - \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\cos \left(2x\right)}{2} \; dx \\\\ & = \dfrac{1}{2} \left[\dfrac{\pi}{2} \; \cos \left(0\right) - 0\right] - \dfrac{1}{4} \left[\sin \left(2x\right)\right]_{0}^{\pi / 2} \\\\ & = \dfrac{- \pi}{4} - \dfrac{1}{4} \left[\sin \left(\pi\right) - \sin \left(0\right)\right] \\\\ & = \dfrac{- \pi}{4} \end{aligned}$

Definite Integration

Evaluate $\displaystyle \int \limits_{0}^{1} \tan^{-1}x \; dx$

Let $I = \displaystyle \int \limits_{0}^{1} \tan^{-1}x \; dx$ $\;\;\; \cdots \; (1)$

Put $\tan^{-1}x = \theta$ $\;\;\; \cdots \; (2)$

Differentiating equation $(2)$ gives

$dx = \sec^2 \theta \; d\theta$ $\;\;\; \cdots \; (2a)$

From equation $(2)$, $x = \tan \theta$

When $x = 0$, $\theta = \tan^{-1}\left(0\right) = 0$ $\;\;\; \cdots \; (2b)$

When $x = 1$, $\theta = \tan^{-1} \left(1\right) = \dfrac{\pi}{4}$ $\;\;\; \cdots \; (2c)$

$\therefore$ $\;$ In view of equations $(2)$, $(2a)$, $(2b)$ and $(2c)$, equation $(1)$ can be written as

$\begin{aligned} I & = \int \limits_{0}^{\frac{\pi}{4}} \theta \; \sec^2 \theta \; d\theta \\\\ & \left[\begin{aligned} \text{Note: }& \int \limits_{a}^{b} u\; v \; dx = \left[u \int v \; dx\right]_{a}^{b} - \int \limits_{a}^{b} \left\{\int v \; dx \times \dfrac{d}{dx} \left(du\right) \right\} \; dx \\\\ & \text{Here } u = \theta; \; v = \sec^2 \theta \end{aligned}\right] \\\\ & = \left[\theta \int \sec^2 \; d\theta\right]_{0}^{\frac{\pi}{4}} - \int \limits_{0}^{\frac{\pi}{4}} \left[\int \sec^2 \theta \; d \theta \times \dfrac{d}{d \theta} \left(\theta\right)\right] d \theta \\\\ & = \left[\theta \; \tan \theta\right]_{0}^{\pi / 4} - \int \limits_{0}^{\frac{\pi}{4}} \tan \theta \; d \theta \\\\ & = \dfrac{\pi}{4} \tan \left(\dfrac{\pi}{4}\right) - 0 - \left[\log \left|\sec \theta\right|\right]_{0}^{\pi / 4} \\\\ & = \dfrac{\pi}{4} - \left[\log \left|\sec \left(\dfrac{\pi}{4}\right)\right| - \log \left|\sec \left(0\right)\right|\right] \\\\ & = \dfrac{\pi}{4} - \left[\log \sqrt{2} - \log 1\right] \\\\ & = \dfrac{\pi}{4} - \log \sqrt{2} = \dfrac{\pi}{4} - \dfrac{1}{2} \log 2 \end{aligned}$

Definite Integration

Evaluate $\displaystyle \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\sin^2 x}{\left(1 + \cos x\right)^2} \; dx$

$\begin{aligned} \text{Let } I & = \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\sin^2 x}{\left(1 + \cos x\right)^2} \; dx \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\left(1 - \cos^2 x\right) \; dx}{\left(1 + \cos x\right)^2} \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\left(1 + \cos x\right) \left(1 - \cos x\right) \; dx}{\left(1 + \cos x\right)^2} \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \dfrac{\left(1 - \cos x\right) \; dx}{1 + \cos x} \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \dfrac{2 \sin^2 \left(\dfrac{x}{2}\right)}{2 \cos^2 \left(\dfrac{x}{2}\right)} \; dx \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \tan^2 \left(\dfrac{x}{2}\right) \; dx \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \left[\sec^2 \left(\dfrac{x}{2}\right) - 1\right] \; dx \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \sec^2 \left(\dfrac{x}{2}\right) \; dx - \int \limits_{0}^{\frac{\pi}{2}} dx \\\\ & = 2 \left[\tan \left(\dfrac{x}{2}\right)\right]_{0}^{\pi / 2} - \left[x\right]_{0}^{\pi / 2} \\\\ & = 2 \left[\tan \left(\dfrac{\pi}{4}\right) - \tan \left(0\right)\right] - \left(\dfrac{\pi}{2} - 0\right) \\\\ & = 2 \left(1 - 0\right) - \dfrac{\pi}{2} = 2 - \dfrac{\pi}{2} \end{aligned}$

Definite Integration

Evaluate $\displaystyle \int \limits_{0}^{\sqrt{2}} \sqrt{2 - x^2} \; dx$

Let $I = \displaystyle \int \limits_{0}^{\sqrt{2}} \sqrt{2 - x^2} \; dx$ $\;\;\; \cdots \; (1)$

Let $x = \sqrt{2} \sin \theta$ $\;\;\; \cdots \; (2)$

Differentiating equation $(2)$ gives

$dx = \sqrt{2} \cos \theta \; d\theta$ $\;\;\; \cdots \; (2a)$

From equation $(2)$, $\theta = \sin^{-1} \left(\dfrac{x}{\sqrt{2}}\right)$

When $x = 0$, $\theta = \sin^{-1} \left(\dfrac{0}{\sqrt{2}}\right) = 0$ $\;\;\; \cdots \; (2b)$

When $x = \sqrt{2}$, $\theta = \sin^{-1} \left(\dfrac{\sqrt{2}}{\sqrt{2}}\right) = \sin^{-1} \left(1\right) = \dfrac{\pi}{2}$ $\;\;\; \cdots \; (2c)$

$\therefore$ $\;$ In view of equations $(2)$, $(2a)$, $(2b)$ and $(2c)$, equation $(1)$ becomes

$\begin{aligned} I & = \int \limits_{0}^{\frac{\pi}{2}} \sqrt{2 - 2 \sin^2 \theta} \times \sqrt{2} \cos \theta \; d\theta \\\\ & = \sqrt{2} \int \limits_{0}^{\frac{\pi}{2}} \sqrt{2 \left(1 - \sin^2 \theta\right)} \cos \theta \; d\theta \\\\ & = 2 \int \limits_{0}^{\frac{\pi}{2}} \cos \theta \times \cos \theta \; d\theta \\\\ & = 2 \int \limits_{0}^{\frac{\pi}{2}} \cos^2 \theta \; d\theta \\\\ & = 2 \int \limits_{0}^{\frac{\pi}{2}} \left(\dfrac{1 + \cos 2 \theta}{2}\right) \; d\theta \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} d\theta + \int \limits_{0}^{\frac{\pi}{2}} \cos 2 \theta \; d\theta \\\\ & = \left[\theta\right]_{0}^{\pi / 2} + \dfrac{1}{2} \left[\sin 2 \theta\right]_{0}^{\pi / 2} \\\\ & = \dfrac{\pi}{2} - 0 + \dfrac{1}{2} \left(\sin \pi - \sin 0\right) \\\\ & = \dfrac{\pi}{2} + \dfrac{1}{2} \left(0 - 0\right) \\\\ & = \dfrac{\pi}{2} \end{aligned}$

Definite Integration

Evaluate $\displaystyle \int \limits_{0}^{\frac{\pi}{2}} \sin^2 x \; dx$

$\begin{aligned} \text{Let } I & = \int \limits_{0}^{\frac{\pi}{2}} \sin^2 x \; dx \\\\ & = \int \limits_{0}^{\frac{\pi}{2}} \left(\dfrac{1 - \cos 2x}{2}\right) \; dx \\\\ & = \dfrac{1}{2} \int \limits_{0}^{\frac{\pi}{2}} dx - \dfrac{1}{2} \int \limits_{0}^{\frac{\pi}{2}} \cos 2x \; dx \\\\ & = \dfrac{1}{2} \left[x\right]_{0}^{\pi / 2} - \dfrac{1}{2} \left[\dfrac{\sin 2x}{2}\right]_{0}^{\pi / 2} \\\\ & = \dfrac{1}{2} \left(\dfrac{\pi}{2} - 0\right) - \dfrac{1}{4} \left(\sin \pi - \sin 0\right) \\\\ & = \dfrac{\pi}{4} - \dfrac{1}{4} \left(0 - 0\right) \\\\ & = \dfrac{\pi}{4} \end{aligned}$

Definite Integration

Obtain $\displaystyle \int \limits_{1}^{3} x^3 \; dx$ as the limits of a sum.

Let $f\left(x\right) = x^3$ $\;\;\; \cdots \; (1)$

$f\left(x\right)$ is continuous on $\left[1, 3\right]$.

Lower limit $= a = 1$; Upper limit $= b = 3$

Divide $\left[1, 3\right]$ into n congruent sub-intervals.

Length of each sub-interval $= h = \dfrac{b - a}{n} = \dfrac{3 - 1}{n} = \dfrac{2}{n}$ $\;\;\; \cdots \; (2)$

$\begin{aligned} \text{Now, } f\left(a + kh\right) & = f\left(1 + kh\right) \\\\ & = \left(1 + k h\right)^3 \;\;\; \left[\text{by equation } (1)\right] \\\\ & = 1 + 3 k h + 3 k^2 h^2 + k^3 h^3 \;\;\; \cdots \; (3) \end{aligned}$

By definition, $\displaystyle \int \limits_{1}^{3} x^3 \; dx = \lim\limits_{n \to \infty} \left[h \sum \limits_{k = 1}^{n} f \left(a + k h\right)\right]$

$\begin{aligned} \therefore \; \int \limits_{1}^{3} x^3 \; dx & = \lim \limits_{n \to \infty} \left[h \sum_{k = 1}^{n} \left(1 + 3 k h + 3 k^2 h^2 + k^3 h^3\right)\right] \;\;\; \left[\text{by equation } (3)\right] \\\\ & = \lim\limits_{n \to \infty} \left[h \sum \limits_{k = 1}^{n} 1 + 3 h \sum \limits_{k = 1}^{n} k + 3 h^2 \sum \limits_{k = 1}^{n} k^2 + h^3 \sum \limits_{k = 1}^{n} k^3\right] \\\\ & = \lim\limits_{n \to \infty} \dfrac{2}{n} \left[n + 3 \times \dfrac{2}{n} \times \dfrac{n \left(n + 1\right)}{2} + 3 \times \left(\dfrac{2}{n}\right)^2 \times \dfrac{n \left(n + 1\right) \left(2 n + 1\right)}{6} \right. \\\\ & \hspace{11em} \left. + \left(\dfrac{2}{n}\right)^3 \times \dfrac{n^2 \left(n + 1\right)^2}{4}\right] \;\;\; \left[\text{by equation } (2)\right] \\\\ & = \lim\limits_{n \to \infty} \dfrac{2}{n} \left[n + 3 \left(n + 1\right) + \dfrac{2 \left(n + 1\right) \left(2 n + 1\right)}{n} + \dfrac{2 \left(n + 1\right)^2}{n}\right] \\\\ & = \lim\limits_{n \to \infty} \dfrac{2}{n} \left[4 n + 3 + \dfrac{2 \left(2 n^2 + 3 n + 1\right)}{n} + \dfrac{2 \left(n^2 + 2 n + 1\right)}{n}\right] \\\\ & = \lim\limits_{n \to \infty} \left[8 + \dfrac{6}{n} + 8 + \dfrac{12}{n} + \dfrac{4}{n^2} + 4 + \dfrac{8}{n} + \dfrac{4}{n^2}\right] \\\\ & \left[\text{As } n \to \infty, \; \dfrac{1}{n} \to 0 \right] \\\\ & = 8 + 8 + 4 = 20 \end{aligned}$

Definite Integration

Obtain $\displaystyle \int \limits_{0}^{\pi} \sin x \; dx$ $\;$ as the limit of a sum.

Let $f\left(x\right) = \sin x$ $\;\;\; \cdots \; (1)$

$f\left(x\right)$ is continuous on $\left[0, \pi\right]$.

Lower limit $= a = 0$; Upper limit $= b = \pi$

Divide $\left[0, \pi\right]$ into n congruent sub-intervals.

Length of each sub-interval $= h = \dfrac{b - a}{n} = \dfrac{\pi - 0}{n} = \dfrac{\pi}{n}$

$\implies$ $n h = \pi$ $\;\;\; \cdots \; (2)$

Now, $f\left(a + kh\right) = f\left(kh\right) = \sin \left(k h\right)$ $\;\;\; \cdots \; (3)$ $\;\;\;$ [from equation $(1)$]

Since $h = \dfrac{1}{n}$ $\implies$ when $n \rightarrow \infty, \; h \rightarrow 0$ $\;\;\; \cdots \; (4)$

By definition, $\displaystyle \int \limits_{0}^{\pi} \sin x \; dx = \lim\limits_{n \to \infty} \left[h \sum \limits_{k = 1}^{n} f \left(a + k h\right)\right]$

$\begin{aligned} \therefore \; \int \limits_{0}^{\pi} \sin x \; dx & = \lim\limits_{h \to 0} \left[h \sum \limits_{k = 1}^{n} \sin\left(k h\right)\right] \;\;\; \left[\text{by equations }(3) \text{ and }(4)\right] \\\\ & = \lim\limits_{h \to 0} h \left[\sin \left(h\right) + \sin \left(2 h\right) + \sin \left(3 h\right) + \cdots + \sin \left(n h\right)\right] \;\;\; \cdots \; (5) \end{aligned}$

Let $S_n = \sin \left(h\right) + \sin \left(2 h\right) + \sin \left(3 h\right) + \cdots + \sin \left(n h\right)$ $\;\;\; \cdots \; (6)$

Multiplying equation $(6)$ with $2 \sin \left(\dfrac{h}{2}\right)$ gives

$\begin{aligned} 2 \sin \left(\dfrac{h}{2}\right) S_n & = 2 \sin \left(h\right) \sin \left(\dfrac{h}{2}\right) + 2 \sin \left(2 h\right) \sin \left(\dfrac{h}{2}\right) \\\\ & \hspace{3em} + 2 \sin \left(3 h\right) \sin \left(\dfrac{h}{2}\right) + \cdots + 2 \sin \left(n h\right) \sin \left(\dfrac{h}{2}\right) \\\\ & \left[\text{Note: } 2 \sin A \; \sin B = \cos \left(A - B\right) - \cos \left(A + B\right)\right] \\\\ & = \cos \left(h - \dfrac{h}{2}\right) - \cos \left(h + \dfrac{h}{2}\right) + \cos \left(2 h - \dfrac{h}{2}\right) - \cos \left(2 h + \dfrac{h}{2}\right) \\\\ & \hspace{3em}+ \cos \left(3 h - \dfrac{h}{2}\right) - \cos \left(3 h + \dfrac{h}{2}\right) + \\\\ & \hspace{7em} \cdots + \cos \left(n h - \dfrac{h}{2}\right) - \cos \left(n h + \dfrac{h}{2}\right) \\\\ & = \cos \left(\dfrac{h}{2}\right) - \cos \left(\dfrac{3 h}{2}\right) + \cos \left(\dfrac{3 h}{2}\right) - \cos \left(\dfrac{5 h}{2}\right) \\\\ & \hspace{3em} + \cos \left(\dfrac{5 h}{2}\right) + \cdots + \cos \left(n h - \dfrac{h}{2}\right) - \cos \left(n h + \dfrac{h}{2}\right) \end{aligned}$

$\begin{aligned} \therefore \; S_n & = \dfrac{\cos \left(\dfrac{h}{2}\right) - \cos \left(n h + \dfrac{h}{2}\right)}{2 \sin \left(\dfrac{h}{2}\right)} \\\\ & = \dfrac{\cos \left(\dfrac{h}{2}\right) - \cos \left(\pi + \dfrac{h}{2}\right)}{2 \sin \left(\dfrac{h}{2}\right)} \;\;\; \left[\text{from equation }(2)\right] \\\\ & = \dfrac{\cos \left(\dfrac{h}{2}\right) + \cos \left(\dfrac{h}{2}\right)}{2 \sin \left(\dfrac{h}{2}\right)} \;\;\; \left[\text{Note: } \cos \left(\pi + \theta\right) = - \cos \theta\right] \\\\ & = \dfrac{2 \cos \left(h / 2\right)}{2 \sin \left(h / 2\right)} \\\\ & = \dfrac{\cos \left(h / 2\right)}{\sin \left(h / 2\right)} \;\;\; \cdots \; (7) \end{aligned}$

$\therefore$ $\;$ In view of equation $(7)$, equation $(5)$ becomes

$\begin{aligned} \int \limits_{0}^{\pi} \sin x \; dx & = \lim\limits_{h \to 0} \; h \times \dfrac{\cos \left(h / 2\right)}{\sin \left(h / 2\right)} \\\\ & = \lim\limits_{\frac{h}{2} \to 0} \; \cos \left(\dfrac{h}{2}\right) \times 2 \times \dfrac{1}{\lim\limits_{\frac{h}{2} \to 0} \; \dfrac{\sin \left(h / 2\right)}{h / 2}} \;\;\; \left[\text{Note: As } h \to 0, \; \dfrac{h}{2} \to 0\right] \\\\ & = 1 \times 2 \times 1 = 2 \end{aligned}$

Definite Integration

Obtain $\displaystyle \int \limits_{\log_{a}2}^{\log_{a}4} a^x \; dx$ as the limit of a sum.

Let $f\left(x\right) = a^x$ $\;\;\; \cdots \; (1)$

$f\left(x\right)$ is continuous on $\left[\log_{a} 2, \log_{a} 4\right]$.

Lower limit $= a = \log_{a} 2$; Upper limit $= b = \log_{a} 4$

Divide $\left[\log_{a} 2, \log_{a} 4\right]$ into n congruent sub-intervals.

Length of each sub-interval $= h = \dfrac{b - a}{n} = \dfrac{\log_{a} 4 - \log_{a} 2}{n} = \dfrac{\log_{a} 2}{n}$

$\implies$ $n h = \log_{a} 2$ $\;\;\; \cdots \; (2)$

Now, $f\left(a + kh\right) = f\left(\left(\log_{a} 2\right) + kh\right) = a^{\left(\log_{a} 2\right) + k h} = a^{\log_{a} 2} \times a^{k h} = 2 a^{k h}$ $\;\;\; \cdots \; (3)$ $\;\;\;$ [from equation $(1)$]

Since $h = \dfrac{1}{n}$ $\implies$ when $n \rightarrow \infty, \; h \rightarrow 0$ $\;\;\; \cdots \; (4)$

By definition, $\displaystyle \int \limits_{\log_{a} 2}^{\log_{a} 4} a^{x} \; dx = \lim\limits_{n \to \infty} \left[h \sum \limits_{k = 1}^{n} f \left(a + k h\right)\right]$

$\begin{aligned} \therefore \; \int \limits_{\log_{a} 2}^{\log_{a} 4} a^x \; dx & = \lim\limits_{h \to 0} \left[h \sum \limits_{k = 1}^{n} 2 a^{k h}\right] \;\;\; \left[\text{by equations }(3) \text{ and }(4)\right] \\\\ & = \lim\limits_{h \to 0} 2 h \left[a^{h} + a^{2 h} + a^{3 h} + \cdots + a^{n h}\right] \\\\ & \left[\begin{aligned} \text{Note: } & a^{h} + a^{2 h} + a^{3 h} + \cdots + a^{n h} \text{ is a geometric series} \\\\ & \text{with first term} = a^{h} \text{ and common ratio } = a^{h} \\\\ & \text{Sum to n terms of a geometric series with first term A} \\\\ & \text{ and common ratio R } = \dfrac{A \left(R^n - 1\right)}{R - 1} \text{ when } R > 1 \end{aligned}\right] \\\\ & = \lim\limits_{h \to 0} 2 h \left[\dfrac{a^{h} \left(a^{n h} - 1\right)}{a^{h} - 1}\right] \\\\ & = \lim\limits_{h \to 0} 2 \; a^{h} \left[\dfrac{a^{\log_{a} 2} - 1}{\dfrac{a^{h} - 1}{h}}\right] \;\;\; \left[\text{by equation } (2)\right] \\\\ & = \dfrac{2 \times a^{0} \times \left(2 - 1\right)}{\log_{e} a} \\\\ & = 2 \times 1 \times 1 \times \log_{a} e \\\\ & = 2 \log_{a} e \end{aligned}$

Definite Integration

Obtain $\displaystyle \int \limits_{0}^{1} e^{2 - 3 x} \; dx$ as the limit of a sum.

Let $f\left(x\right) = e^{2 - 3x}$ $\;\;\; \cdots \; (1)$

$f\left(x\right)$ is continuous on $\left[0,1\right]$.

Lower limit $= a = 0$; Upper limit $= b = 1$

Divide $\left[0,1\right]$ into n congruent sub-intervals.

Length of each sub-interval $= h = \dfrac{b - a}{n} = \dfrac{1 - 0}{n} = \dfrac{1}{n}$

$\implies$ $n h = 1$ $\;\;\; \cdots \; (2)$

Now, $f\left(a + kh\right) = f\left(kh\right) = e^{2 - 3 k h} = \dfrac{e^2}{e^{3 k h}}$ $\;\;\; \cdots \; (3)$ $\;\;\;$ [from equation $(1)$]

Since $h = \dfrac{1}{n}$ $\implies$ when $n \rightarrow \infty, \; h \rightarrow 0$ $\;\;\; \cdots \; (4)$

By definition, $\displaystyle \int \limits_{0}^{1} e^{2 - 3x} \; dx = \lim\limits_{n \to \infty} \left[h \sum \limits_{k = 1}^{n} f \left(a + k h\right)\right]$

$\begin{aligned} \therefore \; \int \limits_{0}^{1} e^{2 - 3x} \; dx & = \lim\limits_{h \to 0} \left[h \sum \limits_{k = 1}^{n} \dfrac{e^2}{e^{3kh}}\right] \;\;\; \left[\text{by equations }(3) \text{ and }(4)\right] \\\\ & = \lim\limits_{h \to 0} e^2 h \left[\dfrac{1}{e^{3h}} + \dfrac{1}{e^{6h}} + \dfrac{1}{e^{9h}} + \cdots + \dfrac{1}{e^{nh}}\right] \\\\ & \left[\begin{aligned} \text{Note: } & \dfrac{1}{e^{3h}} + \dfrac{1}{e^{6h}} + \dfrac{1}{e^{9h}} + \cdots + \dfrac{1}{e^{nh}} \text{ is a geometric series} \\\\ & \text{with first term} = \dfrac{1}{e^{3h}} \text{ and common ratio } = \dfrac{1}{e^{3h}} \\\\ & \text{Sum to n terms of a geometric series with first term A} \\\\ & \text{ and common ratio R } = \dfrac{A \left(1 - R^n\right)}{1 - R} \text{ when } R < 1 \end{aligned}\right] \\\\ & = \lim\limits_{h \to 0} e^2 h \left[\dfrac{\dfrac{1}{e^{3h}} \left(1 - \dfrac{1}{e^{3nh}}\right)}{1 - \dfrac{1}{e^{3h}}}\right] \\\\ & = \lim\limits_{h \to 0} e^2 h \left[\dfrac{1}{e^{3h}} \times \dfrac{\left(1 - \dfrac{1}{e^3}\right)}{e^{3h} - 1} \times e^{3h}\right] \;\;\; \left[\text{by equation } (2)\right] \\\\ & = \dfrac{e^3 -1}{e} \; \lim\limits_{h \to 0} \dfrac{1}{\left(\dfrac{e^{3h} - 1}{3h}\right) \times 3} \\\\ & = \dfrac{1}{3} \left(\dfrac{e^3 - 1}{e}\right) \times \dfrac{1}{\log_{e} e} \\\\ & = \dfrac{1}{3} \left(\dfrac{e^3 - 1}{e}\right) \end{aligned}$

Indefinite Integration

Evaluate $\displaystyle \int \dfrac{1 + \sin x}{\sin x \left(1 + \cos x\right)} \; dx$

$\begin{aligned} \text{Let } I & = \int \dfrac{1 + \sin x}{\sin x \left(1 + \cos x\right)} \; dx \\\\ & = \int \dfrac{dx}{\sin x \left(1 + \cos x\right)} + \int \dfrac{dx}{1 + \cos x} \;\;\; \cdots \; (1) \end{aligned}$

$\begin{aligned} \text{Let } I_1 & = \int \dfrac{dx}{\sin x \left(1 + \cos x\right)} \\\\ & = \int \dfrac{\sin x \; dx}{\sin^2 x \left(1 + \cos x\right)} \\\\ & = \int \dfrac{\sin x \; dx}{\left(1 - \cos^2 x\right) \left(1 + \cos x\right)} \\\\ & = \int \dfrac{\sin x \; dx}{\left(1 - \cos x\right) \left(1 + \cos x\right)^2} \;\;\; \cdots \; (2) \end{aligned}$

Let $\cos x = u$ $\;\;\; \cdots \; (3)$

Differentiating equation $(3)$ gives

$- \sin x \; dx = du$ $\implies$ $\sin x \; dx = - du$ $\;\;\; \cdots \; (3a)$

$\therefore$ $\;$ In view of equations $(3)$ and $(3a)$, equation $(2)$ becomes

$I_1 = \displaystyle \int \dfrac{- du}{\left(1 - u\right) \left(1 + u\right)^2}$ $\;\;\; \cdots \; (4)$

Let $\dfrac{-1}{\left(1 - u\right) \left(1 + u\right)^2} = \dfrac{A}{1 - u} + \dfrac{B}{1 + u} + \dfrac{C}{\left(1 + u\right)^2}$ $\;\;\; \cdots \; (5)$

$\begin{aligned} \implies \; -1 & = A \left(1 + u\right)^2 + B \left(1 - u\right) \left(1 + u\right) + C \left(1 - u\right) \\\\ & = u^2 A + 2 u A + B - u^2 B + C - u C \\\\ & = u^2 \left(A - B\right) + u \left(2 A - C\right) + \left(A + B + C\right) \end{aligned}$

Comparing the coefficients of the $u^2$ term gives

$A - B = 0$ $\implies$ $A = B$ $\;\;\; \cdots \; (5a)$

Comparing the coefficients of the $u$ term gives

$2 A - C = 0$ $\implies$ $C = 2 A$ $\;\;\; \cdots \; (5b)$

Comparing the constant term gives

$-1 = A + B + C$ $\implies$ $-1 = A + A + 2 A$ $\;\;\;$ [by equations $(5a)$ and $(5b)$]

$\implies$ $4 A = -1$ $\implies$ $A = \dfrac{-1}{4}$

$\therefore$ $\;$ From equation $(5a)$, $B = \dfrac{-1}{4}$

From equation $(5b)$, $C = 2 \times \left(\dfrac{-1}{4}\right) = \dfrac{-1}{2}$

Substituting the values of A, B and C in equation $(5)$ gives

$\dfrac{-1}{\left(1 - u\right) \left(1 + u\right)^2} = \dfrac{-1}{4 \left(1 - u\right)} - \dfrac{1}{4 \left(1 + u\right)} - \dfrac{1}{2 \left(1 + u\right)^2}$ $\;\;\; \cdots \; (6)$

$\therefore$ $\;$ We have from equations $(4)$ and $(6)$

$\begin{aligned} I_1 & = \dfrac{-1}{4} \int \dfrac{du}{1 - u} - \dfrac{1}{4} \int \dfrac{du}{1 + u} - \dfrac{1}{2} \int \dfrac{du}{\left(1 + u\right)^2} \\\\ & = \dfrac{1}{4} \log \left|1 - u\right| - \dfrac{1}{4} \log \left|u + 1\right| + \dfrac{1}{2 \left(1 + u\right)} + c_1 \\\\ & = \dfrac{1}{4} \log \left|\dfrac{1 - u}{u + 1}\right| + \dfrac{1}{2 \left(1 + u\right)} + c_1 \\\\ & = \dfrac{1}{4} \log \left|\dfrac{1 - \cos x}{\cos x + 1}\right| + \dfrac{1}{2 \left(1 + \cos x\right)} + c_1 \;\;\; \left[\text{by equation } (3)\right] \\\\ & = \dfrac{1}{4} \log \left|\dfrac{2 \sin^2 \left(\dfrac{x}{2}\right)}{2 \cos^2 \left(\dfrac{x}{2}\right)}\right| + \dfrac{1}{2 \times 2 \cos^2 \left(\dfrac{x}{2}\right)} + c_1 \\\\ & = \dfrac{1}{4} \log \left|\tan^2 \left(\dfrac{x}{2}\right)\right| + \dfrac{1}{4} \sec^2 \left(\dfrac{x}{2}\right) + c_1 \;\;\; \cdots \; (7) \end{aligned}$

$\begin{aligned} \text{Let } I_2 & = \int \dfrac{dx}{1 + \cos x} \\\\ & = \int \dfrac{dx}{2 \cos^2 \left(\dfrac{x}{2}\right)} \\\\ & = \dfrac{1}{2} \int \sec^2 \left(\dfrac{x}{2}\right) \; dx \\\\ & = \dfrac{1}{2} \tan \left(\dfrac{x}{2}\right) \times 2 + c_2 \\\\ & = \tan \left(\dfrac{x}{2}\right) + c_2 \;\;\; \cdots \; (8) \end{aligned}$

$\therefore$ $\;$ In view of equations $(7)$ and $(8)$, equation $(1)$ becomes

$I = \dfrac{1}{4} \log \left|\tan^2 \left(\dfrac{x}{2}\right)\right| + \dfrac{1}{4} \sec^2 \left(\dfrac{x}{2}\right) + \tan \left(\dfrac{x}{2}\right) + c$

where $c = c_1 + c_2$

Indefinite Integration

Evaluate $\displaystyle \int \dfrac{dx}{\sin x \sqrt{\cos^3 x}}$

$\begin{aligned} \text{Let } I & = \int \dfrac{dx}{\sin x \sqrt{\cos^3 x}} \\\\ & = \int \dfrac{\sin x \; dx}{\sin^2 x \sqrt{\cos^3 x}} \\\\ & = \int \dfrac{\sin x \; dx}{\left(1 - \cos^2 x\right) \sqrt{\cos^3 x}} \;\;\; \cdots \; (1) \end{aligned}$

Let $\cos x = u$ $\;\;\; \cdots \; (2)$

Differentiating equation $(2)$ gives

$- \sin x \; dx = du$ $\implies$ $\sin x \; dx = - du$ $\;\;\; \cdots \; (2a)$

$\therefore$ $\;$ In view of equations $(2)$ and $(2a)$, equation $(1)$ becomes

$\begin{aligned} I & = \int \dfrac{- du}{\left(1 - u^2\right) \sqrt{u^3}} \\\\ & = \int \dfrac{- du}{u \sqrt{u} \left(1 + u\right) \left(1 - u\right)} \;\;\; \cdots \; (3) \end{aligned}$

Let $\sqrt{u} = v$ $\;\;\; \cdots \; (4)$

Differentiating equation $(4)$ gives

$\dfrac{1}{2 \sqrt{u}} \; du = dv$ $\implies$ $\dfrac{du}{\sqrt{u}} = 2 \; dv$ $\;\;\; \cdots \; (4a)$

Also from equation $(4)$, $u = v^2$ $\;\;\; \cdots \; (4b)$

$\therefore$ $\;$ In view of equations $(4)$, $(4a)$ and $(4b)$, equation $(3)$ becomes

$I = - 2 \displaystyle \int \dfrac{dv}{v^2 \left(1 + v^2\right) \left(1 - v^2\right)}$ $\;\;\; \cdots \; (5)$

Let $v^2 = t$ (change of variable)

Then, $\dfrac{1}{v^2 \left(1 + v^2\right) \left(1 - v^2\right)} = \dfrac{1}{t \left(1 + t\right) \left(1 - t\right)}$ $\;\;\; \cdots \; (6)$

Let $\dfrac{1}{t \left(1 + t\right) \left(1 - t\right)} = \dfrac{A}{t} + \dfrac{B}{1 + t} + \dfrac{C}{1 - t}$ $\;\;\; \cdots \; (7)$

$\implies$ $1 = A \left(1 + t\right) \left(1 - t\right) + B \; t \left(1 - t\right) + C \; t \left(1 + t\right)$

$\implies$ $1 = A - At^2 + Bt - Bt^2 + Ct + Ct^2$

$\implies$ $1 = t^2 \left(- A - B + C\right) + t \left(B + C\right) + A$

Comparing the constant term gives

$A = 1$ $\;\;\; \cdots \; (7a)$

Comparing the coefficients of the $t$ term gives

$B + C = 0$ $\;\;\; \cdots \; (7b)$

Comparing the coefficients of the $t^2$ term gives

$- A - B + C = 0$ $\implies$ $- B + C = 1$ $\;\;\; \cdots \; (7c)$ $\;\;\;$ [by equation $(7a)$]

Adding equations $(7b)$ and $(7c)$ gives

$2 C = 1$ $\implies$ $C = \dfrac{1}{2}$

$\therefore$ $\;$ From equation $(7b)$, $B = - C = - \dfrac{1}{2}$

Substituting the values of A, B and C in equation $(7)$ gives

$\dfrac{1}{t \left(1 + t\right) \left(1 - t\right)} = \dfrac{1}{t} - \dfrac{1}{2 \left(1 + t\right)} - \dfrac{1}{2 \left(1 - t\right)}$ $\;\;\; \cdots \; (8)$

Since $v^2 = t$, equation $(8)$ becomes

$\dfrac{1}{v^2 \left(1 + v^2\right) \left(1 - v^2\right)} = \dfrac{1}{v^2} - \dfrac{1}{2 \left(1 + v^2\right)} + \dfrac{1}{2 \left(v^2 - 1\right)}$

$\begin{aligned} \therefore \; \int \dfrac{dv}{v^2 \left(1 + v^2\right) \left(1 - v^2\right)} & = \int \dfrac{dv}{v^2} - \dfrac{1}{2} \int \dfrac{dv}{v^2 + 1} + \dfrac{1}{2} \int \dfrac{dv}{v^2 -1} \\\\ & = \dfrac{-1}{v} - \dfrac{1}{2} \tan^{-1} \left(v\right) + \dfrac{1}{2} \int \dfrac{dv}{\left(v + 1\right) \left(v - 1\right)} \;\;\; \cdots \; (9) \end{aligned}$

Let $\dfrac{1}{\left(v + 1\right) \left(v - 1\right)} = \dfrac{P}{v + 1} + \dfrac{Q}{v -1}$ $\;\;\; \cdots \; (10)$

$\implies$ $1 = P \left(v - 1\right) + Q \left(v + 1\right)$

$\implies$ $1 = v \left(P + Q\right) + \left(Q - P\right)$

Comparing the coefficients of the $v$ term gives

$P + Q = 0$ $\implies$ $P = - Q$ $\;\;\; \cdots \; (10a)$

Comparing the constant term gives

$Q - P = 1$ $\implies$ $2 Q = 1$ $\implies$ $Q = \dfrac{1}{2}$ $\;\;\; \cdots \; (10b)$ [by equation $(10a)$]

$\therefore$ $\;$ From equation $(10a)$, $P = \dfrac{-1}{2}$

Substituting the values of P and Q in equation $(10)$ gives

$\dfrac{1}{\left(v + 1\right) \left(v - 1\right)} = \dfrac{-1}{2 \left(v + 1\right)} + \dfrac{1}{2 \left(v - 1\right)}$

$\begin{aligned} \therefore \; \dfrac{1}{2} \int \dfrac{dv}{\left(v + 1\right) \left(v - 1\right)} & = \dfrac{1}{2} \left[\dfrac{-1}{2} \int \dfrac{dv}{v + 1} + \dfrac{1}{2} \int \dfrac{dv}{v - 1}\right] \\\\ & = \dfrac{-1}{4} \log \left|v + 1\right| + \dfrac{1}{4} \log \left|v - 1\right| + c_1 \\\\ & = \dfrac{1}{4} \log \left|\dfrac{v - 1}{v + 1}\right| + c_1 \;\;\; \cdots \; (11) \end{aligned}$

$\therefore$ $\;$ In view of equation $(11)$, equation $(9)$ becomes

$\displaystyle \int \dfrac{dv}{v^2 \left(1 + v^2\right) \left(1 - v^2\right)} = \dfrac{-1}{v} - \dfrac{1}{2} \tan^{-1} \left(v\right) + \dfrac{1}{4} \log \left|\dfrac{v - 1}{v + 1}\right| + c_1$ $\;\;\; \cdots \; (12)$

$\therefore$ $\;$ In view of equation $(12)$, equation $(5)$ becomes

$\begin{aligned} I & = \dfrac{2}{v} + \tan^{-1} \left(v\right) - \dfrac{1}{2} \log \left|\dfrac{v - 1}{v + 1}\right| + c \;\;\; \left[\text{where } c =- 2 c_1\right] \\\\ & = \dfrac{2}{\sqrt{u}} + \tan^{-1} \left(\sqrt{u}\right) - \dfrac{1}{2} \log \left|\dfrac{\sqrt{u} - 1}{\sqrt{u} + 1}\right| + c \;\;\; \left[\text{by equation } (4)\right] \\\\ & = \dfrac{2}{\sqrt{\cos x}} + \tan^{-1} \left(\sqrt{\cos x}\right) - \dfrac{1}{2} \log \left|\dfrac{\sqrt{\cos x} -1 }{\sqrt{\cos x} + 1}\right| + c \;\;\; \left[\text{by equation } (2)\right] \end{aligned}$

Indefinite Integration

Evaluate $\displaystyle \int \left(x - 5\right) \sqrt{x^2 + x} \; dx$

Let $I = \displaystyle \int \left(x - 5\right) \sqrt{x^2 + x} \; dx$ $\;\;\; \cdots \; (1)$

Let $x - 5 = M \; \dfrac{d}{dx} \left(x^2 + x\right) + N$ $\;\;\; \cdots \; (2)$

i.e. $x - 5 = M \left(2 x + 1\right) + N$

i.e. $x - 5 = 2 M x + \left(M + N\right)$

Comparing the coefficients of the $x$ term gives

$1 = 2 M$ $\implies$ $M = \dfrac{1}{2}$ $\;\;\; \cdots \; (3a)$

Comparing the constant term gives

$- 5 = M + N$

$\implies$ $N = - M - 5 = \dfrac{-1}{2} - 5 = \dfrac{-11}{2}$ $\;\;\; \cdots \; (3b)$ $\;\;\; $ [by equation $(3a)$]

In view of equations $(3a)$ and $(3b)$, equation $(2)$ can be written as

$x - 5 = \dfrac{1}{2} \left(2 x + 1\right) - \dfrac{11}{2}$ $\;\;\; \cdots \; (4)$

$\therefore$ $\;$ In view of equation $(4)$, equation $(1)$ becomes

$\begin{aligned} I & = \int \left[\dfrac{1}{2} \left(2x + 1\right) - \dfrac{11}{2}\right] \sqrt{x^2 + x} \; dx \\\\ & = \dfrac{1}{2} \int \left(2 x + 1\right) \sqrt{x^2 + x} \; dx - \dfrac{11}{2} \int \sqrt{x^2 + x} \; dx \;\;\; \cdots \; (5) \end{aligned}$

Let $I_1 = \dfrac{1}{2} \displaystyle \int \left(2 x + 1\right) \sqrt{x^2 + x} \; dx$ $\;\;\; \cdots \; (6)$

Let $x^2 + x = t$ $\;\;\; \cdots \; (6a)$

Differentiating equation $(6a)$ gives

$\left(2 x + 1\right) \; dx = dt$ $\;\;\; \cdots \; (6b)$

$\therefore$ $\;$ In view of equations $(6a)$ and $(6b)$, equation $(6)$ becomes

$\begin{aligned} I_1 & = \dfrac{1}{2} \int \sqrt{t} \; dt \\\\ & = \dfrac{1}{2} \times t^{3/2} \times \dfrac{2}{3} + c_1 \\\\ & = \dfrac{1}{3} t^{3/2} + c_1 \\\\ & = \dfrac{1}{3} \left(x^2 + x\right)^{3/2} + c_1 \;\;\; \cdots \; (7) \;\;\; \left[\text{from equation }(6a)\right] \end{aligned}$

$\begin{aligned} \text{Let } I_2 & = - \dfrac{11}{2} \int \left(2x + 1\right) \sqrt{x^2 + x} \; dx \\\\ & = - \dfrac{11}{2} \int \sqrt{\left(x^2 + x + \dfrac{1}{4}\right) - \dfrac{1}{4}} \; dx \\\\ & = - \dfrac{11}{2} \int \sqrt{\left(x + \dfrac{1}{2}\right)^2 - \left(\dfrac{1}{2}\right)^2} \; dx \\\\ & \left[\text{Note: } \int \sqrt{x^2 - a^2} \; dx = \dfrac{x}{2} \sqrt{x^2 - a^2} - \dfrac{a^2}{2} \log \left|x + \sqrt{x^2 - a^2}\right| + c\right] \\\\ \therefore \; I_2 & = \dfrac{-11}{2} \left(\dfrac{x + \dfrac{1}{2}}{2}\right) \sqrt{\left(x + \dfrac{1}{2}\right)^2 - \left(\dfrac{1}{2}\right)^2} \\ & \hspace{5em} + \dfrac{11}{2} \times \dfrac{\left(1/2\right)^2}{2} \log \left|\left(x + \dfrac{1}{2}\right) + \sqrt{\left(x - \dfrac{1}{2}\right)^2 - \left(\dfrac{1}{2}\right)^2}\right| + c_2 \\\\ & = \dfrac{-11}{2} \left[\left(\dfrac{2x + 1}{4}\right) \sqrt{x^2 + x} - \dfrac{1}{8} \log \left|\dfrac{2x + 1}{2} + \sqrt{x^2 + x}\right|\right] + c_2 \\\\ & = \dfrac{-11}{8} \left(2 x + 1\right) \sqrt{x^2 + x} + \dfrac{11}{16} \log \left|\dfrac{2x + 1}{2} + \sqrt{x^2 + x}\right| + c_2 \;\;\; \cdots \; (8) \end{aligned}$

$\therefore$ $\;$ In view of equations $(7)$ and $(8)$, equation $(5)$ becomes

$I = \dfrac{1}{3} \left(x^2 + x\right)^{3/2} - \dfrac{11}{8} \left(2x + 1\right) \sqrt{x^2 + x} + \dfrac{11}{16} \log \left|\dfrac{2x + 1}{2} + \sqrt{x^2 + x}\right| + c$

where $c = c_1 + c_2$

Indefinite Integration

Evaluate $\displaystyle \int \dfrac{\log x - 1}{\left(\log x\right)^2} \; dx$

$\begin{aligned} \text{Let } I & = \int \dfrac{\log x - 1}{\left(\log x\right)^2} \; dx \\\\ & = \int \left[\dfrac{1}{\log x} - \dfrac{1}{\left(\log x\right)^2}\right] \; dx \;\;\; \cdots \; (1) \end{aligned}$

Let $\log x = t$ $\;\;\; \cdots (2a)$

$\implies$ $x = e^t$ $\;\;\; \cdots \; (2b)$

Differentiating equation $(2a)$ gives

$\dfrac{1}{x} \; dx = dt$ $\implies$ $dx = x \; dt$ $\implies$ $dx = e^t \; dt$ $\;\;\;$ [by equation $(2a)$] $\;\;\; \cdots \; (2c)$

$\therefore$ $\;$ In view of equations $(2a)$ and $(2c)$, equation $(1)$ becomes

$I = \displaystyle \int \left[e^t \left(\dfrac{1}{t} - \dfrac{1}{t^2}\right)\right] \; dt$ $\;\;\; \cdots \; (3)$

If $f\left(t\right) = \dfrac{1}{t}$, then $f'\left(t\right) = \dfrac{-1}{t^2}$

$\therefore$ $\;$ Equation $(3)$ can be rewritten as

$I = \displaystyle \int e^t \left[\dfrac{1}{t} + \dfrac{d}{dt} \left(\dfrac{1}{t}\right)\right] \; dt$

i.e. $I = \dfrac{e^t}{t} + c = \dfrac{e^{\log x}}{\log x} + c$ $\;\;\;$ [by equation $(2a)$]

i.e. $I = \dfrac{x}{\log x} + c$

$\left[\text{Note: }\displaystyle \int e^x \left[f\left(x\right) + f'\left(x\right)\right] \; dx = e^x \; f\left(x\right) + c\right]$

Indefinite Integration

Evaluate $\displaystyle \int \log \left(x + \sqrt{x^2 + a^2}\right) \; dx$

Let $I = \displaystyle \int \log \left(x + \sqrt{x^2 + a^2}\right) \; dx$ $\;\;\; \cdots \; (1)$

Let $x = a \; \tan \theta$ $\;\;\; \cdots \; (2a)$

Differentiating equation $(2a)$ gives

$dx = a \; \sec^2 \theta \; d\theta$ $\;\;\; \cdots \; (2b)$

$\therefore$ $\;$ In view of equations $(2a)$ and $(2b)$, equation $(1)$ becomes

$\begin{aligned} I & = \int \log \left(a \; \tan \theta + \sqrt{a^2 \; \tan^2 \theta + a^2}\right) a \; \sec^2 \theta \; d\theta \\\\ & = a \int \sec^2 \theta \; \log \left(a \; \tan \theta + a \; \sec \theta\right) \; d\theta \\\\ & \left[\begin{aligned} \text{Note: } & \int u \; v \; dx = u \int v \; dx - \int \left\{\int v \; dx \times \dfrac{d}{dx} \left(u\right) \right\} \; dx \\\\ & \text{Here } u = \log \left(a \; \tan \theta + a \; \sec \theta\right), \;\; v = \sec^2 \theta \end{aligned}\right] \\\\ & = a \log \left|a \tan \theta + a \sec \theta\right| \int \sec^2 \theta \; d\theta \\ & \hspace{2em} - a \int \left[\int \sec^2 \theta \; d\theta \times \dfrac{d}{d\theta} \left[\log \left(a \tan \theta + a \sec \theta\right)\right]\right] \; d \theta \\\\ & = a \left\{\tan \theta \; \log \left|a \tan \theta + a \sec \theta\right| - \int \dfrac{\tan \theta \left(a \sec^2 \theta + a \sec \theta \tan \theta\right)}{a \left(\tan \theta + \sec \theta\right)} \; d \theta \right\} \\\\ & = a \left\{\tan \theta \; \log \left|a \tan \theta + a \sec \theta\right| - \int \dfrac{a \tan \theta \sec \theta \left(\sec \theta + \tan \theta\right)}{a \left(\tan \theta + \sec \theta\right)} \; d\theta \right\} \\\\ & = a \tan \theta \; \log \left|a \tan \theta + a \sec \theta\right| - a \int \tan \theta \; \sec \theta \; d\theta \;\;\; \cdots \; (3) \end{aligned}$

$\begin{aligned} \text{Consider } \int \tan \theta \; \sec \theta \; d\theta & = \int \dfrac{\sin \theta}{\cos \theta} \times \dfrac{1}{\cos \theta} \; d \theta \\\\ & = \int \dfrac{\sin \theta}{\cos^2 \theta} \; d \theta \;\;\; \cdots \; (4) \end{aligned}$

Let $\cos \theta = t$ $\;\;\; \cdots \; (5a)$

Differentiating equation $(5a)$ gives

$- \sin \theta \; d\theta = dt$ $\implies$ $\sin \theta \; d\theta = - dt$ $\;\;\; \cdots \; (5b)$

In view of equations $(5a)$ and $(5b)$, equation $(4)$ becomes

$\begin{aligned} \int \tan \theta \; \sec \theta \; d\theta & = \int \dfrac{- dt}{t^2} \\\\ & = \dfrac{1}{t} + c_1 \\\\ & = \dfrac{1}{\cos \theta} + c_1 \;\;\; \left[\text{from equation }(5a)\right] \;\;\; \cdots \; (6) \end{aligned}$

$\therefore$ $\;$ In view of equation $(6)$, equation $(3)$ becomes

$I = a \; \tan \theta \; \log \left|a \tan \theta + a \sec \theta\right| - \dfrac{a}{\cos \theta} + c$ $\;\;\; \cdots \; (7)$

where $c = -a \; c_1$

Now, from equation $(2a)$,

$\tan \theta = \dfrac{x}{a}$ $\;\;\; \cdots \; (8a)$

$\sec \theta = \sqrt{1 + \tan^2 \theta} = \sqrt{1 + \dfrac{x^2}{a^2}} = \dfrac{\sqrt{x^2 + a^2}}{a}$ $\;\;\; \cdots \; (8b)$

Substituting equations $(8a)$ and $(8b)$ in equation $(7)$ gives

$I = a \times \dfrac{x}{a} \times \log \left|a \times \dfrac{x}{a} + a \times \dfrac{\sqrt{x^2 + a^2}}{a}\right| - a \times \dfrac{\sqrt{x^2 + a^2}}{a} + c$

i.e. $I = x \; \log \left|x + \sqrt{x^2 + a^2}\right| - \sqrt{x^2 + a^2} + c$