Indefinite Integration

Evaluate $\int x \; \cos^3 x \; dx$

Let $I = \int x \; \cos^3 x \; dx$ $\;\;\; \cdots$ (1)

$\left[\begin{aligned} \text{Note: } & \cos 3 \theta & = 4 \cos^3 \theta - 3 \cos \theta \\\\ \implies & \cos^3 \theta & = \dfrac{\cos 3 \theta + 3 \cos \theta}{4} \end{aligned}\right]$

$\begin{aligned} \therefore \; I & = \dfrac{1}{4} \int x \left(\cos 3 x + 3 \cos x\right) \; dx \\\\ & = \dfrac{1}{4} \int x \; \cos 3 x \; dx + \dfrac{3}{4} \int x \; \cos x \; dx \\\\ & = \dfrac{1}{4} \; I_1 + \dfrac{3}{4} \; I_2 \;\;\; \cdots (2) \end{aligned}$

$\left[\text{Note: } \displaystyle \int u \cdot v \; dx = u \int v \; dx - \int \left\{\int v \; dx \times \dfrac{d}{dx} \left(u\right) \right\} \; dx \right]$

Consider $I_1 = \displaystyle \int x \; \cos 3 x \; dx$ $\;\;\; \cdots$ (3)

Here $u = x$ and $v = \cos 3 x$

$\begin{aligned} \therefore \; I_1 & = x \int \cos 3 x \; dx - \int \left\{\int \cos 3 x \; dx \times \dfrac{d}{dx} \left(x\right) \right\} \; dx \\\\ & = \dfrac{x \; \sin 3 x}{3} - \int \dfrac{\sin 3 x}{3} \; dx \\\\ & = \dfrac{x \sin 3 x}{3} + \dfrac{\cos 3 x}{9} + c_1 \;\;\; \cdots (3a) \end{aligned}$

Consider $I_2 = \displaystyle \int x \; \cos x \; dx$ $\;\;\; \cdots$ (4)

Here $u = x$ and $v = \cos x$

$\begin{aligned} \therefore \; I_2 & = x \int \cos x \; dx - \int \left\{\int \cos x \; dx \times \dfrac{d}{dx} \left(x\right) \right\} \; dx \\\\ & = x \; \sin x - \int \sin x \; dx \\\\ & = x \; \sin x + \cos x + c_2 \;\;\; \cdots (4b) \end{aligned}$

In view of equations (3a) and (4a), equation (2) becomes

$I = \dfrac{x \; \sin 3 x}{12} + \dfrac{\cos 3 x}{36} + \dfrac{3 \; x \; \sin x}{4} + \dfrac{3 \; \cos x}{4} + c$

where $c = \dfrac{c_1}{4} + \dfrac{3 \; c_2}{4}$

Indefinite Integration

Evaluate $\displaystyle \int \tan^{-1} \left(\dfrac{2 x}{1 - x^2}\right) \; dx$ $\;\;\;$ $0 < x < 1$

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

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

Differentiating equation (2) gives

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

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

$\begin{aligned} \text{Further, } \tan^{-1} \left(\dfrac{2 x}{1 - x^2}\right) & = \tan^{-1} \left(\dfrac{2 \tan \theta}{1 - \tan^2 \theta}\right) \;\;\; \left[\text{In view of equation (2)}\right] \\\\ & = \tan^{-1} \left(\tan 2 \theta\right) \\\\ & = 2 \theta \;\;\; \cdots (2c) \end{aligned}$

In view of equations (2a) and (2c), equation (1) becomes

$I = 2 \int \theta \cdot \sec^2 \theta \; d\theta$ $\;\;\; \cdots$ (3)

$\left[\text{Note: } \displaystyle \int u \cdot v \; dx = u \int v \; dx - \int \left\{\int v \; dx \times \dfrac{d}{dx} \left(u\right) \right\} \; dx \right]$

Here $u = \theta$ and $v = \sec^2 \theta$

$\begin{aligned} \therefore \; I & = 2 \left[\theta \int \sec^2 \theta \; d\theta - \int \left\{\int \sec^2 \theta \; d\theta \times \dfrac{d}{d \theta} \left(\theta\right) \right\} d \theta\right] \\\\ & = 2 \left[\theta \cdot \tan \theta - \int \tan \theta \; d\theta\right] \\\\ & = 2 \; \theta \; \tan \theta - 2 \; \log \left|\sec \theta\right| + c \\\\ & = 2 \; \theta \; \tan \theta - 2 \log \left|\sqrt{1 + \tan^2 \theta}\right| + c \\\\ & = 2 \; \tan^{-1} \left(x\right) \; x - 2 \log \left|\sqrt{1 + x^2}\right| + c \;\;\; \left[\text{In view of equations (2) and (2b)}\right] \\\\ & = 2 \; x \; \tan^{-1} \left(x\right) - \log \left|1 + x^2\right| + c \end{aligned}$

Indefinite Integration

Evaluate $\int \sin \left(\log x\right) \; dx$

Let $I = \int \sin \left(\log x\right) \; dx$ $\;\;\; \cdots$ (1)

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

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

Differentiating equation (2) gives

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

In view of equations (2) and (2b), equation (2a) becomes

$\therefore$ $\;$ $I = \int \sin t \; e^t \; dt$ $\;\;\; \cdots$ (3)

$\left[\text{Note: } \displaystyle \int u \cdot v \; dx = u \int v \; dx - \int \left\{\int v \; dx \times \dfrac{d}{dx} \left(u\right) \right\} \; dx \right]$

Here $u = \sin t$ and $v = e^t$

$\begin{aligned} \therefore \; I & = \sin t \int e^t \; dt - \int \left\{\int e^t \; dt \times \dfrac{d}{dt} \left(\sin t\right) \right\} \; dt \\\\ & = \sin t \cdot e^t - \int e^t \cdot \cos t \; dt \;\;\; \left[\text{Here } u = \cos t \text{ and } v = e^t\right] \\\\ & = e^t \cdot \sin t - \left[\cos t \int e^t \; dt - \int \left\{\int e^t \; dt \times \dfrac{d}{dt} \left(\cos t\right) \right\} \; dt\right] \\\\ & = e^t \cdot \sin t - \left[e^t \cdot \cos t + \int e^t \cdot \sin t \; dt\right] \\\\ & = e^t \cdot \sin t - e^t \cdot \cos t - I + c \;\;\; \left[\text{From equation (3)}\right] \\\\ i.e. \; 2 \; I & = e^t \cdot \sin t - e^t \cdot \cos t + c \\\\ i.e. \; I & = \dfrac{e^t}{2} \left[\sin t - \cos t\right] + c \\\\ & = \dfrac{x}{2} \left[\sin \left(\log x\right) - \cos \left(\log x\right)\right] + c \;\;\; \left[\text{From equations (2) and (2a)}\right] \end{aligned}$

Indefinite Integration

Evaluate $\int \cos^{-1}x \; dx$ $\;\;\;$ $x \in \left[- 1 , 1\right]$

Let $I = \int \cos^{-1}x \; dx = \int 1 \cdot \cos^{-1}x \; dx$

$\left[\text{Note: } \displaystyle \int u \cdot v \; dx = u \int v \; dx - \int \left\{\int v \; dx \times \dfrac{d}{dx} \left(u\right) \right\} \; dx \right]$

Here $u = \cos^{-1} x$ and $v = 1$

$\begin{aligned} \therefore \; I & = \cos^{-1} x \int dx - \int \left\{\int dx \times \dfrac{d}{dx} \left(\cos^{-1}x\right) \right\} \; dx \\\\ & = x \; \cos^{-1} x + \int \dfrac{x}{\sqrt{1 - x^2}} \; dx \;\;\; \cdots (1) \\\\ & \left[\text{Note: } \dfrac{d}{dx} \left(\cos^{-1}x\right) = \dfrac{-1}{\sqrt{1 - x^2}} \right] \end{aligned}$

Consider $\displaystyle \int \dfrac{x}{\sqrt{1 - x^2}} \; dx$ $\;\;\; \cdots$ (2)

Let $1 - x^2 = u$ $\;\;\; \cdots$ (2a)

Differentiating equation (2a) gives

$- 2 x \; dx = du$ $\implies$ $x \; dx = \dfrac{- du}{2}$ $\;\;\; \cdots$ (2b)

In view of equations (2a) and (2b), equation (2) becomes

$\begin{aligned} \int \dfrac{x \; dx}{\sqrt{1 - x^2}} & = \dfrac{-1}{2} \int \dfrac{du}{\sqrt{u}} \\\\ & = \dfrac{-1}{2} \times u^{1/2} \times 2 + c \\\\ & = - \sqrt{u} + c \\\\ & = - \sqrt{1 - x^2} + c \;\;\; \left[\text{From equation (2a)}\right] \;\;\; \cdots (3) \end{aligned}$

In view of equation (3), equation (1) becomes,

$I = x \cos^{-1} x - \sqrt{1 - x^2} + c$

Indefinite Integration

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

Let $I = \displaystyle \int \dfrac{\sin \left(x + a\right)}{\sin \left(x + b\right)} \; dx$ $\;\;\; \cdots$ (1)

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

Differentiating equation (2) gives $dx = du$ $\;\;\; \cdots$ (2a)

Also from equation (2) we have $x = u- b$ $\;\;\; \cdots$ (2b)

In view of equations (2), (2a) and (2b), equation (1) becomes

$\begin{aligned} I & = \int \dfrac{\sin \left(u - b + a\right)}{\sin u} \; du \\\\ & = \int \dfrac{\sin \left[u + \left(a - b\right)\right]}{\sin u} \; du \\\\ & = \int \dfrac{\sin u \cos \left(a - b\right) + \cos u \sin \left(a - b\right)}{\sin u} \; du \\\\ & \left[\text{Note: } \sin \left(\alpha + \beta\right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \right] \\\\ & = \cos \left(a - b\right) \int du + \sin \left(a - b\right) \int \dfrac{\cos u}{\sin u} du \\\\ & = u \cos \left(a - b\right) + \sin \left(a - b\right) \int \cot u \; du \\\\ & = u \cos \left(a - b\right) + \sin \left(a - b\right) \log \left|\sin u\right| + c_1 \\\\ & = \left(x + b\right) \cos \left(a - b\right) + \sin \left(a - b\right) \log \left|\sin \left(x + b\right)\right| + c_1 \;\;\; \left[\text{From equation (2)}\right] \\\\ & = x \cos \left(a - b\right) + \sin \left(a - b\right) \log \left|\sin \left(x + b\right)\right| + c \;\;\; \left[\text{where } c = b \cos \left(a - b\right) + c_1\right] \end{aligned}$

Indefinite Integration

Evaluate $\displaystyle \int \dfrac{x^2 \; dx}{x^2 + 7 x + 10}$ $\;\;\;$ $\left(x \neq - 5, - 2\right)$

Let $I = \displaystyle \int \dfrac{x^2}{x^2 + 7 x + 10} \; dx$ $\;\;\; \cdots$ (1)

$\begin{array}{rll} x^2 + 7 x + 10 ) & x^2 & (1 \\ & \underline{x^2 + 7 x + 10} & \\ & \hspace{6.5mm} - 7 x - 10 & \end{array}$

$\therefore$ $\;$ $\dfrac{x^2}{x^2 + 7 x + 10} = 1 - \dfrac{7 x + 10}{x^2 + 7 x + 10}$ $\;\;\; \cdots$ (2)

Substituting equation (2) in equation (1) gives

$\begin{aligned} I & = \int \left(1 - \dfrac{7 x + 10}{x^2 + 7 x + 10}\right) \; dx \\\\ & = \int dx - \int \dfrac{7 x + 10}{x^2 + 7 x + 10} \; dx \;\;\; \cdots (3) \end{aligned}$

Now, $\int dx = x + c_1$ $\;\;\; \cdots$ (4)

Consider $\displaystyle \int \dfrac{7 x + 10}{x^2 + 7 x + 10} \; dx$ $\;\;\; \cdots$ (5)

Let $7 x + 10 = M \; \dfrac{d}{dx} \left(x^2 + 7 x + 10\right) + N$

i.e. $7 x + 10 = M \left(2 x + 7\right) + N$ $\;\;\; \cdots$ (5a)

i.e. $7 x + 10 = 2 M x + 7 M + N$

Comparing the x coefficients gives

$2 M = 7$ $\implies$ $M = \dfrac{7}{2}$ $\;\;\; \cdots$ (5b)

Comparing the constant terms gives

$10 = 7 M + N$ $\implies$ $N = 10 - 7 M$ $\implies$ $N = 10 - \dfrac{49}{2} = \dfrac{-29}{2}$ $\;\;\; \cdots$ (5c)

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

$\begin{aligned} \int \dfrac{7 x + 10}{x^2 + 7 x + 10} \; dx & = \int \dfrac{\dfrac{7}{2} \left(2 x + 7\right) - \dfrac{29}{2}}{x^2 + 7 x + 10} \; dx \\\\ & = \dfrac{7}{2} \int \dfrac{\left(2 x + 7\right) \; dx}{x^2 + 7 x + 10} - \dfrac{29}{2} \int \dfrac{dx}{x^2 + 7x + 10} \;\;\; \cdots (6) \end{aligned}$

Consider $\displaystyle \int \dfrac{\left(2x + 7\right) \; dx}{x^2 + 7 x + 10}$

Let $x^2 + 7x + 10 = u$

Then, $\left(2 x + 7\right) \; dx = du$

$\begin{aligned} \therefore \; \int \dfrac{\left(2 x + 7\right) \; dx}{x^2 + 7 x + 10} & = \int \dfrac{du}{u} \\\\ & = \log \left|u\right| + c_2 \\\\ & = \log \left|x^2 + 7 x + 10\right| + c_2 \;\;\; \cdots (6a) \end{aligned}$

$\begin{aligned} \text{Consider } \int \dfrac{dx}{x^2 + 7 x + 10} & = \int \dfrac{dx}{\left(x^2 + 7 x + \dfrac{49}{4}\right) + 10 - \dfrac{49}{4}} \\\\ & = \int \dfrac{dx}{\left(x + \dfrac{7}{2}\right)^2 - \left(\dfrac{3}{2}\right)^2} \\\\ & = \dfrac{1}{2} \times \dfrac{2}{3} \log \left|\dfrac{x + \dfrac{7}{2} - \dfrac{3}{2}}{x + \dfrac{7}{2} + \dfrac{3}{2}}\right| + c_3 \\\\ & \left[\text{Note: } \int \dfrac{dx}{x^2 - a^2} = \dfrac{1}{2a} \log \left|\dfrac{x - a}{x + a}\right| + c\right] \\\\ & = \dfrac{1}{3} \log \left|\dfrac{2 x + 4}{2 x + 10}\right| + c_3 \\\\ & = \dfrac{1}{3} \log \left|\dfrac{x + 2}{x + 5}\right| + c_3 \;\;\; \cdots (6b) \end{aligned}$

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

$\displaystyle \int \dfrac{7 x + 10}{x^2 + 7 x + 10} \; dx = \dfrac{7}{2} \log \left|x^2 + 7 x + 10\right| + \dfrac{7}{2} c_2 - \dfrac{29}{6} \log \left|\dfrac{x + 2}{x + 5}\right| - \dfrac{29}{2} c_3$ $\;\;\; \cdots$ (7)

$\therefore$ $\;$ In view of equations (4) and (7), equation (3) becomes

$I = x - \dfrac{7}{2} \log \left|x^2 + 7 x + 10\right| + \dfrac{29}{6} \log \left|\dfrac{x + 2}{x + 5}\right| + c$

where $c = c_1 - \dfrac{7}{2} c_2 + \dfrac{29}{2} c_3$

Indefinite Integration

Evaluate $\displaystyle \int \dfrac{dx}{\left(x + 1\right)^2 \sqrt{x^2 + 2 x + 2}}$

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

Let $\dfrac{1}{x + 1} = p$ $\;\;\; \cdots$ (2a)

Differentiating equation (2a) gives

$- \dfrac{1}{\left(x + 1\right)^2} \; dx = dp$ $\implies$ $\dfrac{dx}{\left(x + 1\right)^2} = - dp$ $\;\;\; \cdots$ (2b)

From equation (2a) we have $x + 1 = \dfrac{1}{p}$ $\;\;\; \cdots$ (2c)

In view of equations (2a), (2b) and (2c), equation (1) becomes

$\begin{aligned} I & = \int \dfrac{-dp}{\sqrt{\dfrac{1}{p^2} + 1}} \\\\ & = - \int \dfrac{p \; dp}{\sqrt{p^2 + 1}} \;\;\; \cdots (3) \end{aligned}$

Let $p^2 + 1 = t$ $\;\;\; \cdots$ (4a)

Differentiating equation (4a) gives

$2 \; p \; dp = dt$ $\implies$ $p \; dp = \dfrac{dt}{2}$ $\;\;\; \cdots$ (4b)

In view of equations (4a) and (4b), equation (3) becomes

$\begin{aligned} I & = - \dfrac{1}{2} \int \dfrac{dt}{\sqrt{t}} \\\\ & = - \dfrac{1}{2} \times 2 \times t^{1/2} + c \\\\ & = - \sqrt{p^2 + 1} + c \;\;\; \left[\text{From equation (4a)}\right] \\\\ & = - \sqrt{\left(\dfrac{1}{x + 1}\right)^2 + 1} + c \;\;\; \left[\text{From equation (2a)}\right] \\\\ & = \dfrac{- \sqrt{x^2 + 2x + 2}}{x + 1} + c \end{aligned}$

Indefinite Integration

Evaluate $\displaystyle \int \dfrac{\sqrt{x}}{\sqrt{x} + \sqrt[3]{x}} \; dx$

Let $I = \displaystyle \int \dfrac{\sqrt{x}}{\sqrt{x} + \sqrt[3]{x}} \; dx$ $\;\;\; \cdots$ (1)

Now, $\dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6}$

Multiply and divide equation (1) with $x^{5/6}$

$\therefore$ $\;$ We have $I = \displaystyle \int \dfrac{1}{x^{5/6}} \times \dfrac{x^{1/2} \times x^{5/6} }{\left(\sqrt{x} + \sqrt[3]{x}\right)} \; dx$

i.e. $I = \displaystyle \int \dfrac{x^{-5/6} \times x^{4/3} \; dx}{x^{1/2} + x^{1/3}}$ $\;\;\; \cdots$ (1a)

Let $x^{1/6} = u$ $\;\;\; \cdots$ (2a)

Differentiating equation (2a) gives

$\dfrac{1}{6} \; x^{-5/6} \; dx = du$ $\implies$ $x^{-5/6} \; dx = 6 du$ $\;\;\; \cdots$ (2b)

From equation (2a),

$x^{1/2} = \left(x^{1/6}\right)^{3} = u^3$ $\;\;\; \cdots$ (2c)

$x^{1/3} = \left(x^{1/6}\right)^{2} = u^2$ $\;\;\; \cdots$ (2d)

$x^{4/3} = \left(x^{1/3}\right)^{4} = \left(u^2\right)^4 = u^8$ $\;\;\; \cdots$ (2e)

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

$\begin{aligned} I & = \int \dfrac{6 \; u^8 \; du}{u^3 + u^2} \\\\ & = \int \dfrac{6 \; u^6 \; du}{u + 1} \;\;\; \cdots (3) \end{aligned}$

$\begin{array}{rll} u + 1 ) & u^6 & (u^5 - u^4 + u^3 - u^2 + u - 1 \\ & \underline{u^6 + u^5} & \\ & \hspace{6mm} -u^5 & \\ & \hspace{6mm} \underline{-u^5 - u^4} & \\ & \hspace{15mm} u^4 & \\ & \hspace{15mm} \underline{u^4 + u^3} & \\ & \hspace{21mm} -u^3 & \\ & \hspace{21mm} \underline{-u^3 - u^2} & \\ & \hspace{30mm} u^2 & \\ & \hspace{30mm} \underline{u^2 + u} & \\ & \hspace{36mm} -u & \\ & \hspace{36mm} \underline{- u - 1} & \\ & \hspace{43.5mm} 1 & \end{array}$

$\therefore$ $\;$ $\dfrac{u^6}{u + 1} = u^5 - u^4 + u^3 - u^2 + u - 1 + \dfrac{1}{u + 1}$ $\;\;\; \cdots$ (4)

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

$\begin{aligned} I & = 6 \left\{\int u^5 \; du - \int u^4 \; du + \int u^3 \; du - \int u^2 \; du + \int u \; du - \int du + \int \dfrac{du}{u + 1} \right\} \\\\ & = 6 \left\{\dfrac{u^6}{6} - \dfrac{u^5}{5} + \dfrac{u^4}{4} - \dfrac{u^3}{3} + \dfrac{u^2}{2} - u + \log \left|u + 1\right| + c \right\} \\\\ & = x - \dfrac{6}{5} \; x^{5/6} + \dfrac{3}{2} \; x^{2/3} - 2 \; x^{1/2} + 3 \; x^{1/3} + 6 \log \left|x^{1/6} + 1\right| + c \\\\ & \left[\text{In view of equation (2a)}\right] \end{aligned}$

Indefinite Integration

Evaluate $\displaystyle \int \dfrac{x^2}{x^4 + x^2 + 1} \; dx$

$\begin{aligned} \text{Let } I & = \int \dfrac{x^2}{x^4 + x^2 + 1} \; dx \\\\ & = \dfrac{1}{2} \int \dfrac{2 x^2}{x^4 + x^2 + 1} \; dx \\\\ & = \dfrac{1}{2} \int \dfrac{\left(x^2 + 1\right) + \left(x^2 - 1\right)}{x^4 + x^2 + 1} \; dx \\\\ & = \dfrac{1}{2} \int \dfrac{x^2 + 1}{x^4 + x^2 + 1} \; dx + \dfrac{1}{2} \int \dfrac{x^2 - 1}{x^4 + x^2 + 1} \; dx \\\\ & = \dfrac{1}{2} \int \dfrac{1 + \dfrac{1}{x^2}}{x^2 + 1 + \dfrac{1}{x^2}} \; dx + \dfrac{1}{2} \int \dfrac{1 - \dfrac{1}{x^2}}{x^2 + 1 + \dfrac{1}{x^2}} \; dx \\\\ & = \dfrac{1}{2} I_1 + \dfrac{1}{2} I_2 \;\;\; \cdots (1) \end{aligned}$

Now, $\left(x - \dfrac{1}{x}\right)^2 = x^2 + \dfrac{1}{x^2} - 2$

$\implies$ $x^2 + \dfrac{1}{x^2} = \left(x - \dfrac{1}{x}\right)^2 + 2$ $\;\;\; \cdots$ (2a)

and $\left(x + \dfrac{1}{x}\right)^2 = x^2 + \dfrac{1}{x^2} + 2$

$\implies$ $x^2 + \dfrac{1}{x^2} = \left(x + \dfrac{1}{x}\right)^2 - 2$ $\;\;\; \cdots$ (2b)

Consider $I_1 = \displaystyle \int \dfrac{\left(1 + \dfrac{1}{x^2}\right) \; dx}{\left(x^2 + \dfrac{1}{x^2}\right) + 1}$ $\;\;\; \cdots$ (3)

In view of equation (2a), equation (3) can be written as

$\begin{aligned} I_1 & = \displaystyle \int \dfrac{\left(1 + \dfrac{1}{x^2}\right) \; dx}{\left(x - \dfrac{1}{x}\right)^2 + 2 + 1} \\\\ & = \displaystyle \int \dfrac{\left(1 + \dfrac{1}{x^2}\right) \; dx}{\left(x - \dfrac{1}{x}\right)^2 + 3} \;\;\; \cdots (3a) \end{aligned}$

In equation (3a), let $x - \dfrac{1}{x} = u$ $\;\;\; \cdots$ (4a)

Differentiating equation (4a) gives

$\left(1 + \dfrac{1}{x^2}\right) \; dx = du$ $\;\;\; \cdots$ (4b)

In view of equations (4a) and (4b), equation (3a) becomes,

$\begin{aligned} I_1 & = \int \dfrac{du}{u^2 + 3} \\\\ & = \int \dfrac{du}{\left(u\right)^2 + \left(\sqrt{3}\right)^2} \\\\ & = \dfrac{1}{\sqrt{3}} \tan^{-1} \left(\dfrac{u}{\sqrt{3}}\right) + c_1 \;\;\; \left[\text{Note: } \int \dfrac{dx}{x^2 + a^2} = \dfrac{1}{a} \tan^{-1} \left(\dfrac{x}{a}\right ) + c\right] \\\\ & = \dfrac{1}{\sqrt{3}} \tan^{-1} \left(\dfrac{x - \dfrac{1}{x}}{\sqrt{3}}\right) + c_1 \;\;\; \left[\text{From equation (4a)}\right] \\\\ & = \dfrac{1}{\sqrt{3}} \tan^{-1} \left(\dfrac{x^2 - 1}{x \sqrt{3}}\right) + c_1 \;\;\; \cdots (3b) \end{aligned}$

Consider $I_2 = \displaystyle \int \dfrac{\left(1 - \dfrac{1}{x^2}\right) \; dx}{\left(x^2 + \dfrac{1}{x^2}\right) + 1}$ $\;\;\; \cdots$ (5)

In view of equation (2b), equation (5) can be written as

$\begin{aligned} I_2 & = \int \dfrac{\left(1 - \dfrac{1}{x^2}\right) \; dx}{\left(x + \dfrac{1}{x}\right)^2 - 2 + 1} \\\\ & = \int \dfrac{\left(1 - \dfrac{1}{x^2}\right) \; dx}{\left(x + \dfrac{1}{x}\right)^2 - 1} \;\;\; \cdots (5a) \end{aligned}$

In equation (5a), let $x + \dfrac{1}{x} = v$ $\;\;\; \cdots$ (6a)

Differentiating equation (6a) gives

$\left(1 - \dfrac{1}{x^2}\right) \; dx = dv$ $\;\;\; \cdots$ (6b)

In view of equations (6a) and (6b), equation (5a) becomes,

$\begin{aligned} I_2 & = \int \dfrac{dv}{v^2 - 1} \\\\ & = \int \dfrac{dv}{\left(v\right)^2 - \left(1\right)^2} \\\\ & = \dfrac{1}{2} \log \left|\dfrac{v - 1}{v + 1}\right| + c_2 \;\;\; \left[\text{Note: } \int \dfrac{dx}{x^2 - a^2} = \dfrac{1}{2a} \log \left|\dfrac{x - a}{x + a}\right| + c\right] \\\\ & = \dfrac{1}{2} \log \left|\dfrac{x + \dfrac{1}{x} - 1}{x + \dfrac{1}{x} + 1}\right| + c_2 \;\;\; \left[\text{From equation (6a)}\right] \\\\ & = \dfrac{1}{2} \log \left|\dfrac{x^2 - x + 1}{x^2 + x + 1}\right| + c_2 \;\;\; \cdots (5b) \end{aligned}$

Putting equations (3b) and (5b) in equation (1) gives

$I = \dfrac{1}{2 \sqrt{3}} \tan^{-1} \left(\dfrac{x^2 - 1}{x \sqrt{3}}\right) + \dfrac{1}{4} \log \left|\dfrac{x^2 - x +1}{x^2 + x + 1}\right| + c$

where $c = \dfrac{c_1}{2} + \dfrac{c_2}{2}$

Indefinite Integration

Evaluate $\displaystyle \int \dfrac{x^2 + 1}{x^4 + 1} \; dx$

$\begin{aligned} \text{Let } I & = \displaystyle \int \dfrac{x^2 + 1}{x^4 + 1} \; dx \\\\ & = \displaystyle \int \dfrac{1 + \dfrac{1}{x^2}}{x^2 + \dfrac{1}{x^2}} \; dx \;\;\; \cdots (1) \end{aligned}$

Now, $\left(x - \dfrac{1}{x}\right)^2 = x^2 - 2 + \dfrac{1}{x^2}$

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

In view of equation (2), equation (1) becomes

$I = \displaystyle \int \dfrac{1 + \dfrac{1}{x^2}}{\left(x - \dfrac{1}{x}\right)^2 + 2} \; dx$ $\;\;\;\; \cdots$ (3)

Let $x - \dfrac{1}{x} = t$ $\;\;\; \cdots$ (4a)

Differentiating equation (4a) gives,

$\left(1 + \dfrac{1}{x^2}\right) \; dx = dt$ $\;\;\; \cdots$ (4b)

Substituting equations (4a) and (4b) in equation (3) gives

$\begin{aligned} I & = \int \dfrac{dt}{t^2 + 2} \\\\ & = \int \dfrac{dt}{\left(t\right)^2 + \left(\sqrt{2}\right)^2} \\\\ & = \dfrac{1}{\sqrt{2}} \; \tan^{-1} \left(\dfrac{t}{\sqrt{2}}\right) + c \;\;\; \left[\text{Note: } \int \dfrac{dx}{x^2 + a^2} = \dfrac{1}{a} \tan^{-1} \left(\dfrac{x}{a}\right) + c\right] \\\\ & = \dfrac{1}{\sqrt{2}} \tan^{-1} \left(\dfrac{x - \dfrac{1}{x}}{\sqrt{2}}\right) + c \;\;\; \left[\text{From equation (4a)}\right] \\\\ & = \dfrac{1}{\sqrt{2}} \tan^{-1} \left(\dfrac{x^2 - 1}{x \sqrt{2}}\right) + c \end{aligned}$

Indefinite Integration

Evaluate $\displaystyle \int \dfrac{2 \sin 2 x - \cos x}{6 - \cos^2 x - 4 \sin x} \; dx$

$\begin{aligned} \text{Let } I & = \int \dfrac{2 \sin 2 x - \cos x}{6 - \cos^2 x - 4 \sin x} \; dx \\\\ & = \int \dfrac{4 \sin x \cos x - \cos x}{6 - \left(1 - \sin^2 x\right) - 4 \sin x} \; dx \;\;\; \left[\text{Note: } \sin 2 x = 2 \sin x \cos x\right] \\\\ & = \int \dfrac{\cos x \left(4 \sin x - 1\right)}{\sin^2 x - 4 \sin x + 5} \; dx \;\;\; \cdots (1) \end{aligned}$

In equation (1), let $\sin x = t$ $\;\;\; \cdots$ (2a)

Differentiating equation (2a) gives

$\cos x \; dx = dt$ $\;\;\; \cdots$ (2b)

In view of equations (2a) and (2b), equation (1) becomes

$I = \displaystyle \int \dfrac{\left(4 t - 1\right) \; dt}{t^2 - 4 t + 5}$ $\;\;\; \cdots$ (3)

In equation (3), let

$4 t - 1 = P \dfrac{d}{dt} \left(t^2 - 4 t + 5\right) + Q$

i.e. $4 t - 1 = P \left(2 t - 4\right) + Q$ $\;\;\; \cdots$ (4a)

i.e. $4 t - 1 = 2 P t - 4 P + Q$

Comparing the coefficients of t and the constant term on both sides gives

$2 P = 4$ $\implies$ $P = 2$

and $- 1 = -4 P + Q$ $\implies$ $Q = 4 P - 1 = 7$

Substituting the values of P and Q in equation (4a) gives

$4 t - 1 = 2 \left(2 t - 4\right) + 7$ $\;\;\; \cdots$ (4b)

In view of equation (4b), equation (3) becomes

$\begin{aligned} I & = \int \dfrac{2 \left(2 t - 4\right) + 7}{t^2 - 4 t + 5} \; dt \\\\ & = \int \dfrac{4 t - 8}{t^2 - 4 t + 5} \; dt + 7 \int \dfrac{dt}{t^2 - 4 t + 5} \\\\ & = I_1 + 7 I_2 \;\;\; \cdots (5) \end{aligned}$

Consider $I_1 = \displaystyle \int \dfrac{4 t - 8}{t^2 - 4 t + 5} \; dt$ $\;\;\; \cdots$ (5a)

In equation (5a), let $t^2 - 4 t + 5 = z$ $\;\;\; \cdots$ (6a)

Differentiating equation (6a) gives

$\left(2 t - 4\right) dt = dz$

$\therefore$ $\left(4 t - 8\right) dt = 2 dz$ $\;\;\; \cdots$ (6b)

In view of equations (6a) and (6b), equation (5a) becomes

$\begin{aligned} I_1 & = \int \dfrac{2 \; dz}{2} \\\\ & = 2 \; \log \left|z\right| + c_1 \\\\ & = 2 \; \log \left|t^2 - 4 t + 5\right| + c_1 \;\;\; \cdots (7a) \end{aligned}$

Consider

$\begin{aligned} I_2 & = \int \dfrac{dt}{t^2 - 4 t + 5} \;\;\; \cdots (5b) \\\\ & = \int \dfrac{dt}{\left(t^2 - 4 t + 4\right) + 5 - 4} \\\\ & = \int \dfrac{dt}{\left(t - 2\right)^2 + \left(1\right)^2} \\\\ & = \tan^{-1} \left(t - 2\right) + c_2 \;\;\; \cdots (7b) \;\;\; \left[\text{Note: } \int \dfrac{dx}{x^2 + a^2} = \dfrac{1}{a} \tan^{-1} \left(\dfrac{x}{a}\right) + c\right] \end{aligned}$

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

$\begin{aligned} I & = 2 \log \left|t^2 - 4 t + 5\right| + c_1 + 7 \tan^{-1} \left(t - 2\right) + c_2 \\\\ & = 2 \log \left|\sin^2 x - 4 \sin x + 5\right| + 7 \tan^{-1} \left(\sin x - 2\right) + c \;\;\; \left[\text{from equation (2a)}\right] \end{aligned}$

where $c = c_1 + c_2$

Indefinite Integration

Evaluate $\displaystyle \int \dfrac{3 x + 2}{2 x^2 + x + 1} \; dx$

Let $I = \displaystyle \int \dfrac{3 x + 2}{2 x^2 + x + 1} \; dx$ $\;\;\; \cdots$ (1)

Let $3 x + 2 = m \dfrac{d}{dx} \left(2 x^2 + x + 1\right) + n$

i.e. $3 x + 2 = m \left(4 x + 1\right) + n$ $\;\;\; \cdots$ (2)

i.e. $3 x + 2 = 4 m x = m + n$

Comparing the coefficient of x and constant term on both sides, we get

$4 m = 3$ $\implies$ $m = \dfrac{3}{4}$ $\;\;\; \cdots$ (3a)

and $m + n = 2$ $\implies$ $n = 2 - m = 2 - \dfrac{3}{4} = \dfrac{5}{4}$ $\;\;\; \cdots$ (3b)

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

$3 x + 2 = \dfrac{3}{4} \left(4 x + 1\right) + \dfrac{5}{4}$ $\;\;\; \cdots$ (4)

In view of equation (4), equation (1) becomes

$\begin{aligned} I & = \int \dfrac{\dfrac{3}{4} \left(4 x + 1\right) + \dfrac{5}{4}}{2 x^2 + x + 1} \; dx \\\\ & = \dfrac{3}{4} \int \dfrac{4 x + 1}{2 x^2 + x + 1} \; dx + \dfrac{5}{4} \int \dfrac{dx}{2 x^2 + x + 1} \\\\ & = \dfrac{3}{4} \; I_1 + \dfrac{5}{4} \; I_2 \;\;\; \cdots (5) \end{aligned}$

where $I_1 = \displaystyle \int \dfrac{4x + 1}{2 x^2 + x + 1} \; dx$ $\;\;\; \cdots$ (5a)

and $I_2 = \displaystyle \int \dfrac{dx}{2 x^2 + x + 1} \; dx$ $\;\;\; \cdots$ (5b)

Consider $I_1$. Put $2 x^2 + x + 1 = p$ $\;\;\; \cdots$ (6a)

Differentiating equation (6a) gives $\left(4 x + 1\right) \; dx = dp$ $\;\;\; \cdots$ (6b)

In view of equations (6a) and (6b), equation (5a) becomes

$\begin{aligned} I_1 & = \displaystyle \int \dfrac{dp}{p} \\\\ & = \log \left|p\right| + c_1 \\\\ & = \log \left|2 x^2 + x + 1\right| + c_1 \;\;\; \cdots (7a) \;\;\; \left[\text{In view of equation (6a)}\right] \end{aligned}$

$\begin{aligned} I_2 & = \int \dfrac{dx}{2 x^2 + x + 1} \\\\ & = \dfrac{1}{2} \int \dfrac{dx}{x^2 + \dfrac{x}{2} + \dfrac{1}{2}} \\\\ & = \dfrac{1}{2} \int \dfrac{dx}{\left(x^2 + \dfrac{1}{2} x + \dfrac{1}{16}\right) + \dfrac{1}{2} - \dfrac{1}{16}} \\\\ & = \dfrac{1}{2} \int \dfrac{dx}{\left(x + \dfrac{1}{4}\right)^2 + \left(\dfrac{\sqrt{7}}{4}\right)^2} \\\\ & \left[\text{Note: } \int \dfrac{dx}{x^2 + a^2} = \dfrac{1}{a} \tan^{-1} \left(\dfrac{x}{a}\right)\right] \\\\ & = \dfrac{1}{2} \times \dfrac{4}{\sqrt{7}} \tan^{-1} \left(\dfrac{x + \dfrac{1}{4}}{\dfrac{\sqrt{7}}{4}}\right) + c_2 \\\\ & = \dfrac{2}{\sqrt{7}} \tan^{-1} \left(\dfrac{4 x + 1}{\sqrt{7}}\right) + c_2 \;\;\; \cdots (7b) \end{aligned}$

$\therefore$ We have from equations (7a), (7b) and (5)

$\begin{aligned} I & = \dfrac{3}{4} \log \left|2 x^2 + x + 1\right| + c_1 + \dfrac{5}{4} \times \dfrac{2}{\sqrt{7}} \tan^{-1} \left(\dfrac{4 x + 1}{\sqrt{7}}\right) + c_2 \\\\ & = \dfrac{3}{4} \log \left|2 x^2 + x + 1\right| + \dfrac{5}{2 \sqrt{7}} \tan^{-1} \left(\dfrac{4 x + 1}{\sqrt{7}}\right) + c \end{aligned}$

where $c = c_1 + c_2$

Indefinite Integration

Evaluate $\displaystyle \int \dfrac{dx}{x^2 + 3 x + 3}$

$\begin{aligned} \text{Let } I & = \int \dfrac{dx}{x^2 + 3 x + 3} \\\\ & = \int \dfrac{dx}{\left(x^2 + 3 x + \dfrac{9}{4}\right) + 3 - \dfrac{9}{4}} \\\\ & \left[\text{Note: Completion of square in the denominator}\right] \\\\ & = \int \dfrac{dx}{\left(x + \dfrac{3}{2}\right)^2 + \left(\dfrac{\sqrt{3}}{2}\right)^2} \\\\ & = \dfrac{2}{\sqrt{3}} \tan^{-1} \left(\dfrac{x + \dfrac{3}{2}}{\dfrac{\sqrt{3}}{2}}\right) + c \;\;\; \left[\text{Note: } \int \dfrac{dx}{x^2 + a^2} = \dfrac{1}{a} \tan^{-1} \left(\dfrac{x}{a}\right)\right] \\\\ & = \dfrac{2}{\sqrt{3}} \tan^{-1} \left(\dfrac{2 x + 3}{\sqrt{3}}\right) + c \end{aligned}$

Indefinite Integration

Evaluate $\int \sin^4 x \; \cos^2 x \; dx$

Let $I = \int \sin^4 x \; \cos^2 x \; dx$ $\;\;\; \cdots$ (1)

Compare equation (1) with $\int \sin^m x \; \cos^n x \; dx$

Here $m = 4$ (even) and $n = 2$ (even)

Now,

$\begin{aligned} \sin^4 x \; \cos^2 x & = \dfrac{1}{4} \times \left(4 \; \sin^2 x \; \cos^2 x\right) \times \sin^2 x \\\\ & = \dfrac{1}{4} \times \left(2 \; \sin x \; \cos x\right)^2 \times \sin^2 x \\\\ & = \dfrac{1}{4} \times \sin^2 2x \times \sin^2 x \\\\ & = \dfrac{1}{4} \; \left(\dfrac{1 - \cos 4x}{2}\right) \; \left(\dfrac{1 - \cos 2 x}{2}\right) \;\;\; \left[\text{Note: } \sin^2 \theta = \dfrac{1 - \cos 2 \theta}{2} \right] \\\\ & = \dfrac{1}{16} \; \left(1 - \cos 2 x - \cos 4 x + \cos 2x \; \cos 4x\right) \\\\ & = \dfrac{1}{16} \; \left(1 - \cos 2 x - \cos 4 x + \dfrac{2 \cos 2 x \; \cos 4 x}{2}\right) \\\\ & \left[\text{Note: } 2 \cos C \; \cos D = \cos \left(C + D\right) + \cos \left(C - D\right) \right] \\\\ & = \dfrac{1}{32} \; \left(2 - 2 \cos 2 x - 2 \cos 4 x + \cos 6 x + \cos 2 x\right) \\\\ & = \dfrac{1}{32} \; \left(\cos 6 x - 2 \cos 4 x - \cos 2 x + 2\right) \;\;\; \cdots (2) \end{aligned}$

In view of equation (2), equation (1) becomes

$\begin{aligned} I & = \dfrac{1}{32} \int \left(\cos 6 x - 2 \cos 4 x - \cos 2 x + 2\right) \; dx \\\\ & = \dfrac{1}{32} \; \left\{\int \cos 6 x \; dx - 2 \int \cos 4 x \; dx - \int \cos 2 x \; dx + 2 \int dx \right\} \\\\ & = \dfrac{1}{32} \; \left\{\dfrac{\sin 6 x}{6} - \dfrac{\sin 4 x}{2} - \dfrac{\sin 2 x}{2} + 2 x \right\} + c \end{aligned}$

Indefinite Integration

Evaluate $\int \sin^5 x \; \cos^4 x \; dx$

Let $I = \int \sin^5 x \; \cos^4 x \; dx $ $\;\;\; \cdots$ (1)

Compare equation (1) with $\int \sin^m x \; \cos^n x \; dx$

Here $m = 5$ (odd) and $n = 4$ (even)

$\therefore$ $\;$ Let $\cos x = p$ $\;\;\; \cdots$ (2a)

Differentiating equation (2a) gives

$- \sin x \; dx = dp$ $\;\;\; \cdots$ (2b)

Now, equation (1) can be rewritten as

$\begin{aligned} I & = \int \sin x \cdot \sin^4 x \cdot \cos^4 x \; dx \\\\ & = \int \sin x \cdot \left(\sin^2 x\right)^2 \cdot \cos^4 x \; dx \\\\ & = \int \sin x \cdot \left(1 - \cos^2 x\right)^2 \cdot \cos^4 x \; dx \\\\ & = - \int \left(1 - p^2\right)^2 \cdot p^4 \; dp \;\;\; \left[\text{In view of equations (2a) and (2b)}\right] \\\\ & = - \int \left(1 - 2 p^2 + p^4\right) \cdot p^4 \; dp \\\\ & = \int \left(- p^4 + 2 p^6 - p^8 \right) \; dp \\\\ & = - \int p^4 \; dp + 2 \int p^6 \; dp - \int p^8 \; dp \\\\ & = - \dfrac{p^5}{5} + \dfrac{2 \; p^7}{7} - \dfrac{p^9}{p^9} + c \\\\ & = \dfrac{- \cos^5 x}{5} + \dfrac{2 \cos^7 x}{7} - \dfrac{\cos^9 x}{9} + c \;\;\; \left[\text{In view of equation (2a)}\right] \end{aligned}$

Indefinite Integration

Evaluate $\displaystyle \int \dfrac{dx}{\cos x - \sin x}$ $\;\;\;$ $\left(0 < x < \dfrac{\pi}{4}\right)$

Let $I = \displaystyle \int \dfrac{dx}{\cos x - \sin x}$ $\;\;\; \cdots$ (1)

Put $\tan \left(\dfrac{x}{2}\right) = t$ $\;\;\; \cdots$ (2a)

Then, $\dfrac{1}{2} \sec^2 \left(\dfrac{x}{2}\right) dx = dt$

i.e. $dx = \dfrac{2 \; dt}{\sec^2 \left(\dfrac{x}{2}\right)} = \dfrac{2 \; dt}{1 + \tan^2 \left(\dfrac{x}{2}\right)} = \dfrac{2 \; dt}{1 + t^2}$ $\;\;\; \cdots$ (2b)

Also, $\cos x = \dfrac{1 - \tan^2 \left(\dfrac{x}{2}\right)}{1 + \tan^2 \left(\dfrac{x}{2}\right)} = \dfrac{1 - t^2}{1 + t^2}$ $\;\;\; \cdots$ (2c)

and $\sin x = \dfrac{2 \tan \left(\dfrac{x}{2}\right)}{1 + \tan^2 \left(\dfrac{x}{2}\right)} = \dfrac{2 \; t}{1 + t^2}$ $\;\;\; \cdots$ (2d)

In view of equations (2a), (2b), (2c) and (2d), equation (1) becomes

$\begin{aligned} I & = \int \dfrac{\dfrac{2 \; dt}{1 + t^2}}{\dfrac{1 - t^2}{1 + t^2} - \dfrac{2 \; t}{1 + t^2}} \\\\ & = \int \dfrac{2 \; dt}{1 - t^2 - 2 t} \\\\ & = 2 \int \dfrac{dt}{2 - \left(t^2 + 2 t + 1\right)} \\\\ & = 2 \int \dfrac{dt}{\left(\sqrt{2}\right)^2 - \left(t + 1\right)^2} \\\\ & \left[\text{Note: } \int \dfrac{dx}{a^2 - x^2} = \dfrac{1}{2 a} \log \left|\dfrac{x + a}{x - a}\right| + c \right] \\\\ & = 2 \times \dfrac{1}{2 \sqrt{2}} \log \left|\dfrac{t + 1 + \sqrt{2}}{t - 1 - \sqrt{2}}\right| + c \\\\ & = \dfrac{1}{\sqrt{2}} \log \left|\dfrac{\tan \left(\dfrac{x}{2}\right) + 1 + \sqrt{2}}{\tan \left(\dfrac{x}{2}\right) - 1 - \sqrt{2} }\right| + c \;\;\; \left[\text{From equation (2a)}\right] \end{aligned}$

Indefinite Integration

Evaluate $\displaystyle \int \dfrac{dx}{3 + 2 \sin x + \cos x}$

Let $I = \displaystyle \int \dfrac{dx}{3 + 2 \sin x + \cos x}$ $\;\;\; \cdots$ (1)

Put $\tan \left(\dfrac{x}{2}\right) = t$ $\;\;\; \cdots$ (2a)

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

$\implies$ $dx = \dfrac{2 \; dt}{\sec^2 \left(\dfrac{x}{2}\right)} = \dfrac{2 \; dt}{1 + \tan^2 \left(\dfrac{x}{2}\right)} = \dfrac{2 \; dt}{1 + t^2}$ $\;\;\; \cdots$ (2b)

Also, $\sin x = \dfrac{2 \; \tan \left(\dfrac{x}{2}\right)}{1 + \tan^2 \left(\dfrac{x}{2}\right)} = \dfrac{2 \; t}{1 + t^2}$ $\;\;\; \cdots$ (2c)

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

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

$\begin{aligned} I & = 2 \displaystyle \int \dfrac{dt}{\left(1 + t^2\right) \left(3 + \dfrac{4t}{1 + t^2} + \dfrac{1 - t^2}{1 + t^2}\right)} \\\\ & = 2 \int \dfrac{dt}{3 + 3 t^2 + 4 t + 1 - t^2} \\\\ & = 2 \int \dfrac{dt}{2 t^2 + 4 t + 4} \\\\ & = \int \dfrac{dt}{t^2 + 2 t + 2} \\\\ & = \int \dfrac{dt}{\left(t^2 + 2 t + 1\right) + 1} \\\\ & = \int \dfrac{dt}{\left(t + 1\right)^2 + \left(1\right)^2} \\\\ & = \tan^{-1} \left(t + 1\right) + c \;\;\; \cdots (3) \;\; \left[\text{Note: } \int \dfrac{dx}{x ^2 + a^2} = \dfrac{1}{a} \tan^{-1} \left(\dfrac{x}{a} + c\right) \right] \end{aligned}$

Substituting the value of t from equation (2a) in equation (3) gives

$I = \tan^{-1} \left[\tan \left(\dfrac{x}{2}\right) + 1 \right] + c$

Indefinite Integration

Evaluate $\displaystyle\int \dfrac{x}{\left(16 - 9 x^2\right)^{3/2}} \; dx$ $\;\;$ $\left(0 < x < \dfrac{4}{3}\right)$

$\begin{aligned} \text{Let } I & = \int \dfrac{x}{\left(16 - 9 x^2\right)^{3/2}} \; dx \\\\ & = \int \dfrac{x}{\left\{\sqrt{9 \left(\dfrac{16}{9} - x^2\right)} \right\}^3} \; dx \\\\ & = \dfrac{1}{27} \int \dfrac{x \; dx}{\left\{\sqrt{\left(\dfrac{4}{3}\right)^2 - \left(x\right)^2} \right\}^3} \;\;\; \cdots (1) \end{aligned}$

Put $x = \dfrac{4}{3} \sin \theta$ $\;\;\; \cdots$ (2a)

Then, $dx = \dfrac{4}{3} \cos \theta \; d \theta$ $\;\;\; \cdots$ (2b)

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

$\begin{aligned} I & = \dfrac{1}{27} \int \dfrac{\dfrac{4}{3} \; \sin \theta \times \dfrac{4}{3} \; \cos \theta \; d \theta}{\left\{\sqrt{\left(\dfrac{4}{3}\right)^2 - \left(\dfrac{4}{3} \sin \theta\right)^2} \right\}^3} \\\\ & = \dfrac{1}{27} \times \dfrac{16}{9} \int \dfrac{\sin \theta \; \cos \theta \; d \theta}{\left\{\sqrt{\dfrac{16}{9} \left(1 - \sin^2 \theta\right)} \right\}^3} \\\\ & = \dfrac{1}{27} \times \dfrac{16}{9} \int \dfrac{\sin \theta \; \cos \theta \; d \theta}{\left(\dfrac{4}{3} \; \cos \theta\right)^3} \\\\ & = \dfrac{1}{27} \times \dfrac{16}{9} \times \dfrac{27}{64} \int \dfrac{\sin \theta \; \cos \theta \; d \theta}{\cos^3 \theta} \\\\ & = \dfrac{1}{36} \int \dfrac{\sin \theta \; d \theta}{\cos^2 \theta} \\\\ & = \dfrac{1}{36} \int \dfrac{1}{\cos \theta} \times \dfrac{\sin \theta}{\cos \theta} \; d \theta \\\\ & = \dfrac{1}{36} \int \sec \theta \; \tan \theta \; d \theta \\\\ & = \dfrac{1}{36} \sec \theta + c \;\;\; \cdots (3) \end{aligned}$

From equation (2a),

$\sin \theta = \dfrac{3 x}{4}$

$\therefore$ $\;$ $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \dfrac{9 x^2}{16}} = \dfrac{\sqrt{16 - 9 x^2}}{4}$

$\therefore$ $\sec \theta = \dfrac{4}{\sqrt{16 - 9x^2}}$ $\;\;\; \cdots$ (4)

Substituting the value of $\sec \theta$ from equation (4) in equation (3) gives

$\begin{aligned} I & = \dfrac{1}{36} \times \dfrac{4}{\sqrt{16 - 9 x^2}} + c \\\\ & = \dfrac{1}{9} \times \dfrac{1}{\sqrt{16 - 9 x^2}} + c \end{aligned}$

Indefinite Integration

Evaluate $\displaystyle\int x^2 \sqrt{a^6 - x^6} \; dx$

Let $I = \displaystyle\int x^2 \sqrt{a^6 - x^6} \; dx$ $\;\;\; \cdots$ (1)

Put $x^3 = u$ $\;\;\; \cdots$ (2a) $\implies$ $x^6 = u^2$ $\;\;\; \cdots$ (2b)

Differentiating equation (2a) gives

$3 x^2 dx = du$ $\implies$ $x^2 dx = \dfrac{du}{3}$ $\;\;\; \cdots$ (3)

$\therefore$ $\;$ In view of equations (2b) and (3), equation (1) becomes

$I = \dfrac{1}{3} \displaystyle\int \sqrt{a^6 - u^2} \; du$ $\;\;\; \cdots$ (4)

Put $u = a^3 \sin \theta$ $\;\;\; \cdots$ (5a) $\implies$ $u^3 = a^6 \sin^2 \theta$ $\;\;\; \cdots$ (5b)

Differentiating equation (5a) gives, $du = a^3 \cos \theta \; d\theta$ $\;\;\; \cdots$ (6)

$\therefore$ In view of equations (5b) and (6), equation (4) becomes

$\begin{aligned} I & = \dfrac{1}{3} \int a^3 \cos \theta \sqrt{a^6 - a^6 \sin^2 \theta} \; d\theta \\\\ & = \dfrac{a^3}{3} \int \cos \theta \sqrt{a^6 \left(1 - \sin^2 \theta\right)} \; d\theta \\\\ & = \dfrac{a^6}{3} \int \cos^2 \theta \; d\theta \\\\ & = \dfrac{a^6}{3} \int \dfrac{1 + \cos 2 \theta}{2} \; d\theta \\\\ & = \dfrac{a^6}{6} \left(\int d \theta + \int \cos 2 \theta \; d \theta\right) \\\\ & = \dfrac{a^6}{6} \left(\theta + \dfrac{\sin 2 \theta}{2}\right) + c \\\\ & = \dfrac{a^6}{6} \left(\theta + \dfrac{2 \sin \theta \cos \theta}{2}\right) + c \\\\ & = \dfrac{a^6}{6} \left(\theta + \sin \theta \cos \theta\right) + c \;\;\; \cdots (7) \end{aligned}$

Now, from equation (5a) we have

$\sin \theta = \dfrac{u}{a^3}$ $\;\;\; \cdots$ (8a) $\implies$ $\theta = \sin^{-1} \left(\dfrac{u}{a^3}\right)$ $\;\;\; \cdots$ (8b)

Also, $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \dfrac{u^2}{a^6}} = \dfrac{\sqrt{a^6 - u^2}}{a^3}$ $\;\;\; \cdots$ (8c)

$\therefore$ $\;$ In view of equations (8a), (8b) and (8c), equation (7) can be written as

$\begin{aligned} I & = \dfrac{a^6}{6} \left\{\sin^{-1} \left(\dfrac{u}{a^3}\right) + \dfrac{u}{a^3} \times \dfrac{\sqrt{a^6 - u^2}}{a^3} \right\} + c \\\\ & = \dfrac{a^6}{6} \sin^{-1} \left(\dfrac{u}{a^3}\right) + \dfrac{u \sqrt{a^6 - u^2}}{6} + c \;\;\; \cdots (9) \end{aligned}$

In view of equation (2a), equation (9) becomes

$I = \dfrac{a^6}{6} \sin^{-1} \left(\dfrac{x^3}{a^3}\right) + \dfrac{x^3 \sqrt{a^6 - x^6}}{6}$ + c

Indefinite Integration

Evaluate $\displaystyle\int \dfrac{dx}{\sqrt{x + 1} + \sqrt[3]{x + 1}}$

Let $I = \displaystyle\int \dfrac{dx}{\sqrt{x + 1} + \sqrt[3]{x + 1}}$

Put $x + 1 = p^6$ $\;\;\; \cdots$ (1)

Then, $dx = 6 p^5 dp$

Further, $\sqrt{x + 1} = \left(p^6\right)^{1/2} = p^3$ and $\sqrt[3]{x + 1} = \left(p^6\right)^{1/3} = p^2$

$\begin{aligned} \therefore \; I & = \int \dfrac{6 \; p^5 \; dp}{p^3 + p^2} \\\\ & = 6 \int \dfrac{p^5 \; dp}{p^2 \left(p + 1\right)} \\\\ & = 6 \int \dfrac{p^3 \; dp}{p + 1} \end{aligned}$

Put $p + 1 = t$ $\implies$ $p = t - 1$ $\;\;\; \cdots$ (2)

Differentiating equation (2) gives $dp = dt$

Also, from equation (2), $p^3 = \left(t - 1\right)^3 = t^3 - 3 t^2 + 3 t - 1$

$\begin{aligned} \therefore \; I & = 6 \int \dfrac{t^3 - 3 t^2 + 3 t - 1}{t} \; dt \\\\ & = 6 \left\{\int t^2 \; dt - 3 \int t \; dt + 3 \int dt - \int \dfrac{dt}{t} \right\} \\\\ & = 6 \left\{\dfrac{t^3}{3} - \dfrac{3 t^2}{2} + 3 t - \log \left|t\right| \right\} + k \\\\ & = 2 t^3 - 9 t^2 + 18 t - 6 \log \left|t\right| + k \\\\ & = 2 \left(p + 1\right)^3 - 9 \left(p + 1\right)^2 + 18 \left(p + 1\right) - 6 \log \left|p + 1\right| + k \;\;\; \cdots (3) \end{aligned}$

Now, from equation (1),

$p = \left(x + 1\right)^{1/6}$ $\;\;\; \cdots$ (4)

$\begin{aligned} \therefore \; \left(p + 1\right)^3 & = \left[\left(x + 1\right)^{1/6} + 1\right]^3 \\\\ & = \left[\left(x + 1\right)^{1/6}\right]^3 + 3 \left[\left(x + 1\right)^{1/6}\right]^2 + 3 \left(x + 1\right)^{1/6} + 1 \\\\ & = \left(x + 1\right)^{1/2} + 3 \left(x + 1\right)^{1/3} + 3 \left(x + 1\right)^{1/6} + 1 \;\;\; \cdots (5) \end{aligned}$

$\begin{aligned} \left(p + 1\right)^2 & = \left[\left(x + 1\right)^{1/6} + 1\right]^2 \\\\ & = \left[\left(x + 1\right)^{1/6}\right]^2 + 2 \left(x + 1\right)^{1/6} + 1 \\\\ & = \left(x + 1\right)^{1/3} + 2 \left(x + 1\right)^{1/6} + 1 \;\;\; \cdots (6) \end{aligned}$

Substituting equations (4), (5) and (6) in equation (3) gives

$\begin{aligned} I & = 2 \left[\left(x + 1\right)^{1/2} + 3 \left(x + 1\right)^{1/3} + 3 \left(x + 1\right)^{1/6} + 1\right] \\\\ & - 9 \left[\left(x + 1\right)^{1/3} + 2 \left(x + 1\right)^{1/6} + 1\right] + 18 \left[\left(x + 1\right)^{1/6} + 1\right] - 6 \log \left|\left(x + 1\right)^{1/6} + 1\right| + k \\\\ & = 2 \left(x + 1\right)^{1/2} - 3 \left(x + 1\right)^{1/3} + 6 \left(x + 1\right)^{1/6} - 6 \log \left|\left(x + 1\right)^{1/6} + 1\right| + c \end{aligned}$

where $11 + k = c = \text{constant}$

Indefinite Integration

Evaluate $\displaystyle\int \dfrac{\sin x}{\sin 3x} \; dx$

$\begin{aligned} \text{Let } I & = \int \dfrac{\sin x}{\sin 3x} \; dx \\\\ & = \int \dfrac{\sin x}{3 \sin x - 4 \sin^3 x} \; dx \;\;\; \left[\text{Note: } \sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta\right] \\\\ & = \int \dfrac{\sin x}{3 \sin x \left(1 - \sin^2 x\right) - \sin^3 x} \; dx \\\\ & = \int \dfrac{\sin x}{3 \sin x \cos^2 x - \sin^3 x} \; dx \end{aligned}$

Now, $\sin x = \dfrac{\sin x / \cos x}{1 / \cos x} = \dfrac{\tan x}{\sec x}$ and $\cos x = \dfrac{1}{\sec x}$

$\begin{aligned} \therefore \; 3 \sin x \cos^2 x - \sin^3 x & = 3 \times \dfrac{\tan x}{\sec x} \times \dfrac{1}{\sec^2 x} - \dfrac{\tan^3 x}{\sec^3 x} \\\\ & = \dfrac{\tan x \left(3 - \tan^2 x\right)}{\sec^3 x} \end{aligned}$

$\begin{aligned} \therefore \; I & = \int \dfrac{\tan x / \sec x}{\tan x \left(3 - \tan^2 x\right) / \sec^3 x} \; dx \\\\ & = \int \dfrac{\sec^2 x}{3 - \tan^2 x} \; dx \end{aligned}$

Let $\tan x = p$

Then, $\sec^2 x \; dx = dp$

$\begin{aligned} \therefore \; I & = \int \dfrac{dp}{3 - p^2} \\\\ & = \int \dfrac{dp}{\left(\sqrt{3}\right)^2 - \left(p\right)^2} \\\\ & = \dfrac{1}{2 \sqrt{3}} \log \left|\dfrac{\sqrt{3} + p}{\sqrt{3} - p}\right| + c \;\;\; \left[\text{Note: } \int \dfrac{dx}{a^2 - x^2} = \dfrac{1}{2a} \log \left|\dfrac{x + a}{x - a}\right| \right] \\\\ & = \dfrac{1}{2 \sqrt{3}} \log \left|\dfrac{\sqrt{3} + \tan x}{\sqrt{3} - \tan x}\right| + c \end{aligned}$

Indefinite Integration

Evaluate $\displaystyle\int \dfrac{dx}{\sin\left(x - a\right) \sin \left(x - b\right)}$

$\begin{aligned} \text{Let } I & = \int \dfrac{dx}{\sin \left(x-a\right) \sin \left(x - b\right)} \\\\ & = \dfrac{1}{\sin \left(b - a\right)} \int \dfrac{\sin \left(b - a\right)}{\sin \left(x - a\right) \sin \left(x - b\right)} \; dx \\\\ & = \dfrac{1}{\sin \left(b - a\right)} \int \dfrac{\sin \left[\left(x - a\right) - \left(x - b\right)\right]}{\sin \left(x - a\right) \sin \left(x - b\right)} \; dx \\\\ & = \dfrac{1}{\sin \left(b - a\right)} \int \dfrac{\sin \left(x - a\right) \cos \left(x - b\right) - \cos \left(x - a\right) \sin \left(x - b\right)}{\sin \left(x - a\right) \sin \left(x - b\right)} \; dx \\\\ & \left[\text{Note: } \sin \left(\alpha - \beta\right) = \sin \alpha \cos \beta - \cos \alpha \sin \beta\right] \\\\ & = \dfrac{1}{\sin \left(b - a\right)} \int \left[\int \dfrac{\cos \left(x - b\right)}{\sin \left(x - b\right)} \; dx - \int \dfrac{\cos \left(x - a\right)}{\sin \left(x - a\right)} \; dx\right] \\\\ & = \dfrac{1}{\sin \left(b - a\right)} \left[\int \cot \left(x - b\right) \; dx - \int \cot \left(x - a\right) \; dx\right] \\\\ & = \dfrac{1}{\sin \left(b - a\right)} \left[\log \left|\sin \left(x - b\right)\right| - \log \left|\sin \left(x - a\right)\right|\right] + c \\\\ & = \dfrac{1}{\sin \left(b - a\right)} \log \left|\dfrac{\sin \left(x - b\right)}{\sin \left(x - a\right)}\right| + c \end{aligned}$

Indefinite Integration

Evaluate $\int \cos 2x \; \cos 4 x \; \cos 6x \; dx$

$\begin{aligned} \text{Let } I & = \int \cos 2x \; \cos 4x \; \cos 6x \; dx \\\\ & = \dfrac{1}{2} \int 2 \cos 2x \; \cos 4x \; \cos 6x \; dx \\\\ & = \dfrac{1}{2} \int \left\{\cos \left(4x + 2x\right) + \cos \left(4x - 2x\right) \right\} \cos 6x \; dx \\\\ & = \dfrac{1}{2} \int \left(\cos 6x + \cos 2x\right) \cos 6x \; dx \\\\ & = \dfrac{1}{2} \int \cos^2 6x \; dx + \dfrac{1}{2} \int \cos 2x \; \cos 6x \; dx \\\\ & = \dfrac{1}{2} \int \dfrac{1 + \cos 12x}{2} \; dx + \dfrac{1}{4} \int 2 \; \cos 2x \; \cos 6x \; dx \\\\ & = \dfrac{1}{4} \int dx + \dfrac{1}{4} \int \cos 12x \; dx + \dfrac{1}{4} \int \cos \left(6x+2x\right) + \cos \left(6x - 2x\right) \; dx \\\\ & = \dfrac{1}{4} \left\{\int dx + \int \cos 12x \; dx + \int \cos 8x \; dx + \int \cos 4x \; dx \right\} \\\\ & = \dfrac{1}{4} \left\{x + \dfrac{\sin 12x}{12} + \dfrac{\sin 8x}{8} + \dfrac{\sin 4x}{4} \right\} + c \\\\ & = \dfrac{x}{4} + \dfrac{\sin 12x}{48} + \dfrac{\sin 8x}{32} + \dfrac{\sin 4x}{16} + c \\\\ & \left[\begin{aligned} \text{Note: } & 2 \cos A \; \cos B = \cos \left(A + B\right) + \cos \left(A - B\right) \\\\ & \cos^2 \theta = \dfrac{1 + \cos 2 \theta}{2} \end{aligned}\right] \end{aligned}$

Indefinite Integration

Evaluate $\displaystyle\int \dfrac{\sin 4x}{\sin x} \; dx$

$\begin{aligned} \text{Let } I & = \int \dfrac{\sin 4x}{\sin x} \; dx \\\\ & = \int \dfrac{\sin\left[2\left(2x\right)\right]}{\sin x} \; dx \\\\ & = \int \dfrac{2 \sin 2x \cos 2x}{\sin x} \; dx \;\;\; \left[\text{Note: } \sin 2 \theta = 2 \sin \theta \cos \theta\right] \\\\ & = \int \dfrac{4 \sin x \cos x \cos 2x}{\sin x} \; dx \\\\ & = 4 \int \cos x \cos 2x \; dx \\\\ & = 4 \int \cos x \left(1 - 2 \sin^2 x\right) \; dx \;\; \left[\text{Note: } \cos 2x = 1 - 2 \sin^2 x\right] \\\\ & = 4 \int \cos x \; dx - 8 \int \cos x \; \sin^2 x \; dx \;\;\; \cdots (1) \end{aligned}$

Consider $\int \cos x \; \sin^2 x \; dx$

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

Differentiating equation (2) gives $\cos x \; dx = du$

$\begin{aligned} \therefore \; \int \cos x \; \sin^2 x \; dx & = \int u^2 \; du \\\\ & = \dfrac{u^3}{3} + c_1 \\\\ & = \dfrac{\sin^3 x}{3} + c_1 \;\;\; \cdots (3) \end{aligned}$

Also, $\int \cos x \; dx = \sin x + c_2$ $\;\;\; \cdots$ (4)

Substituting equations (3) and (4) in equation (1) gives

$I = 4 \sin x - \dfrac{8}{3} \sin^3 x + c$ where $c = c_1 + c_2$

Indefinite Integration

Evaluate $\displaystyle\int \dfrac{8x + 13}{\sqrt{4x + 7}} \; dx$

Let $I = \displaystyle\int \dfrac{8x + 3}{\sqrt{4x + 7}} \; dx$

Let $4x + 7 = u$ $\;\;\; \cdots$ (1) $\implies$ $x = \dfrac{u - 7}{4}$

Differentiating equation (1) gives

$4 dx = du$ $\implies$ $dx = \dfrac{du}{4}$

Now, $8x + 13 = 8 \left(\dfrac{u - 7}{4}\right) + 13 = 2 u - 1$

$\begin{aligned} \therefore \; I & = \dfrac{1}{4} \int \dfrac{2 u - 1}{\sqrt{u}} \; du \\\\ & = \dfrac{1}{2} \times u^{3/2} \times \dfrac{2}{3} - \dfrac{1}{4} \times u^{1/2} \times 2 + c \\\\ & = \dfrac{1}{3} \left(4x + 7\right)^{3/2} - \dfrac{1}{2} \left(4x + 7\right)^{1/2} + c \end{aligned}$

Indefinite Integration

Evaluate $\displaystyle\int \dfrac{\cot x}{\text{cosec }x - \cot x} dx$

$\begin{aligned} \text{Let } I & = \int \dfrac{\cot x}{\text{cosec }x - \cot x} dx \\\\ & = \int \dfrac{\cot x \left(\text{cosec }x + \cot x\right)}{\left(\text{cosec }x - \cot x\right) \left(\text{cosec }x + \cot x\right)} dx \\\\ & = \int \dfrac{\cot x \;\; \text{cosec }x + \cot^2 x}{\text{cosec}^2 x - \cot^2 x} dx \\\\ & = \int \cot x \;\; \text{cosec }x \;\; dx + \int \cot^2 x \;\; dx \;\;\;\; \left[\text{Note: }\text{cosec}^2 x - \cot^2 x = 1\right] \end{aligned}$

$\begin{aligned} \text{Now, } \int \cot^2 x \;\; dx & = \int \dfrac{\cos^2 x}{\sin^2 x} \;\; dx \\\\ & = \int \dfrac{1 - \sin^2 x}{\sin^2 x} \;\; dx \\\\ & = \int \dfrac{1}{\sin^2 x} \;\; dx - \int dx \\\\ & = \int \text{cosec}^2 x \;\; dx - \int dx \end{aligned}$

$\begin{aligned} \therefore I & = \int \cot x \;\; \text{cosec }x \;\; dx + \int \text{cosec}^2 x \;\; dx - \int dx \\\\ & = - \text{cosec }x - \cot x - x + c \end{aligned}$

Indefinite Integration

Evaluate $\displaystyle\int \dfrac{\cos x}{\cos x - 1} dx$

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

Indefinite Integration

Evaluate $\displaystyle\int \dfrac{x^6 + 2}{x^2 + 1} dx$

Let $I = \displaystyle\int \dfrac{x^6 + 1}{x^2 + 2} dx$

i.e. $I = \displaystyle\int \dfrac{x^6 + 1}{x^2 + 1} dx \;\; + \;\; \displaystyle\int \dfrac{1}{x^2 + 1} dx$

i.e. $I = I_1 + I_2$ where $I_1 = \displaystyle\int \dfrac{x^6 + 1}{x^2 + 1} dx$ and $I_2 = \displaystyle\int \dfrac{1}{x^2 + 1} dx$

Now, $\dfrac{x^6 + 1}{x^2 + 1} = \dfrac{\left(x^2 + 1\right) \left(x^4 - x^2 + 1\right)}{x^2 + 1} = x^4 - x^2 + 1$

$\therefore$ $I_1 = \int \left(x^4 - x^2 + 1\right) dx$

i.e. $I_1 = \int x^4 dx - \int x^2 dx + \int dx$

$\therefore$ $I = \int x^4 dx - \int x^2 dx + \int dx + \displaystyle\int \dfrac{1}{x^2 + 1} dx$

i.e. $I = \dfrac{x^5}{5} - \dfrac{x^3}{3} + x + \tan^{-1} x + c$

Application of Derivatives: Increasing Decreasing Functions

Find the intervals in which $f\left(x\right) = \dfrac{x}{\log x}$ is increasing or decreasing.

$f\left(x\right) = \dfrac{x}{\log x}$

$\therefore$ $f'\left(x\right) = \dfrac{\log x - x \times \dfrac{1}{x}}{\left(\log x\right)^2}$

i.e. $f'\left(x\right) = \dfrac{\log x -1}{\left(\log x\right)^2}$

Now, $\left(\log x\right)^2 > 0 \;\; \forall \; x $

$\therefore$ $f'\left(x\right) > 0$ $\implies$ $\log x - 1 > 0$

i.e. $\log x > 1$ $\implies$ $\log x > \log e$ $\implies$ $x>e$

$\therefore$ $f\left(x\right)$ is increasing in the interval $\left(e,\infty\right)$

$f'\left(x\right) < 0$ $\implies$ $\log x -1 < 0$

i.e. $\log x < 1$ $\implies$ $x < e$

$\therefore$ $f\left(x\right)$ is decreasing in the interval $\left(0,e\right)$

Application of Derivatives: Increasing Decreasing Functions

Determine the intervals in which $f:R \rightarrow R$, $f\left(x\right) = 2 \cos x + \sin^2 x$ is strictly increasing or strictly decreasing.

$f\left(x\right) = 2 \cos x + \sin^2 x$

$\therefore$ $f'\left(x\right) = -2 \sin x + 2 \sin x \cos x$

i.e. $f'\left(x\right) = -2 \sin x \left(1 - \cos x\right) $

i.e. $f'\left(x\right) = -2 \sin x \times 2 \sin^2 \left(\dfrac{x}{2}\right)$

i.e. $f'\left(x\right) = -4 \sin x \sin^2 \left(\dfrac{x}{2}\right)$ $\;\; \cdots$ (1)

Now, $\sin^2 \left(\dfrac{x}{2}\right) > 0 \;\; \forall \; x \in R$

When $f'\left(x\right) > 0$ $\implies$ $f\left(x\right)$ is increasing.

Now, from equation (1), $f'\left(x\right) > 0$ $\implies$ $\sin x < 0$

$\implies$ $- \pi < x < 0$

Generalizing, we have $\left(2 k - 1\right)\pi < x < 2 k \pi$

i.e. $f\left(x\right)$ is strictly increasing in the interval $\left(\left(2 k - 1\right)\pi, 2 k \pi\right), \;\; k \in Z$

When $f'\left(x\right) < 0$ $\implies$ $f\left(x\right)$ is decreasing.

Now, from equation (1), $f'\left(x\right) < 0$ $\implies$ $\sin x > 0$

$\implies$ $0 < x < \pi$

Generalizing, we have $2 k \pi < x < \left(2 k + 1\right) \pi$

i.e. $f\left(x\right)$ is strictly decreasing in the interval $\left(2 k \pi, \left(2 k + 1 \right) \pi \right), \;\; k \in Z$

Application of Derivatives: Increasing Decreasing Functions

Find the intervals in which the function $f\left(x\right) = \dfrac{4x^2 + 1}{x}$ is increasing and the intervals in which it is decreasing.

$f\left(x\right) = \dfrac{4x^2 + 1}{x}$ $\implies$ $f\left(x\right) = 4x + \dfrac{1}{x}$

$\therefore$ $f'\left(x\right) = 4 - \dfrac{1}{x^2} = \dfrac{4x^2 -1}{x^2}$

Now, $x^2 > 0 \;\; \forall \; x, \;\; x \neq 0$

$f'\left(x\right) = 0$ $\implies$ $4x^2 -1 = 0$

i.e. $x^2 = \dfrac{1}{4}$ $\implies$ $x = \dfrac{1}{2}$, $x = -\dfrac{1}{2}$

These points divide the real line into three intervals namely $\left(-\infty, -\dfrac{1}{2}\right)$, $\left(-\dfrac{1}{2}, \dfrac{1}{2}\right)$ and $\left(\dfrac{1}{2},\infty\right)$

Interval	Sign of $\left(4x^2 - 1\right)$	Sign of $f'\left(x\right)$	Nature of function f
$\left(-\infty,-\dfrac{1}{2}\right)$	+ve	+ve	Increasing
$\left(-\dfrac{1}{2},\dfrac{1}{2}\right)$	-ve	-ve	Decreasing
$\left(\dfrac{1}{2},\infty\right)$	+ve	+ve	Increasing

$\therefore$ $f\left(x\right)$ is increasing in the interval $\left\{x : x < \dfrac{-1}{2}\right\} \cup \left\{x : x > \dfrac{1}{2}\right\} $

$f\left(x\right)$ is decreasing in the interval $\left\{x : \dfrac{-1}{2} < x < \dfrac{1}{2}\right\} $

Application of Derivatives: Increasing Decreasing Functions

Determine the intervals in which $f\left(x\right) = \sin^4 x + \cos^4 x$, $x \in \left(0,\dfrac{\pi}{2}\right)$ is increasing and in which it is decreasing.

$f\left(x\right) = \sin^4 x + \cos^4 x$

$\therefore$ $f'\left(x\right) = 4 \sin^3 x \cos x - 4 \cos^3 x \sin x$

i.e. $f'\left(x\right) = 4 \sin x \cos x \left(\sin^2 x - \cos^2 x\right)$

i.e. $f'\left(x\right) = -4 \sin x \cos x \cos 2x$ $\;\; \cdots$ (1)

Now, $\sin x > 0 \; \forall \; x \in \left(0,\dfrac{\pi}{2}\right)$; $\cos x > 0 \; \forall \; x \in \left(0,\dfrac{\pi}{2}\right)$

$f\left(x\right)$ is increasing when $f'\left(x\right) > 0$

Since $\sin x > 0$ and $\cos x > 0$, $\therefore$ $\;$ $f'\left(x\right) > 0$ $\implies$ $\cos 2x < 0$

i.e. $\dfrac{\pi}{2} < 2x < \pi$ or $\pi < 2x < \dfrac{3\pi}{2}$

i.e. $\dfrac{\pi}{4} < x < \dfrac{\pi}{2}$ or $\dfrac{\pi}{2} < x < \dfrac{3\pi}{4}$

Since $x \in \left(0,\dfrac{\pi}{2}\right)$, therefore $f\left(x\right)$ is increasing in $\left(\dfrac{\pi}{4}, \dfrac{\pi}{2}\right)$

$f\left(x\right)$ is decreasing when $f'\left(x\right) < 0$

Since $\sin x > 0$ and $\cos x > 0$, $\therefore$ $\;$ $f'\left(x\right) < 0$ $\implies$ $\cos 2x > 0$

i.e. $0 < 2x < \dfrac{\pi}{2}$ or $\dfrac{3\pi}{2} < 2x < 2\pi$

i.e. $0 < x < \dfrac{\pi}{4}$ or $\dfrac{3\pi}{4} < x < \pi $

Since $x \in \left(0,\dfrac{\pi}{2}\right)$, therefore $f\left(x\right)$ is decreasing in $\left(0,\dfrac{\pi}{4}\right)$

Application of Derivatives: Increasing Decreasing Functions

Find the intervals in which the function $f\left(x\right) = \sin x - \cos x$, $0 < x < 2 \pi$ is increasing or decreasing.

$f\left(x\right) = \sin x - \cos x$

$\begin{aligned} \therefore \; f'\left(x\right) & = \cos x + \sin x \\ & = \sqrt{2}\left(\dfrac{1}{\sqrt{2}} \cos x + \dfrac{1}{\sqrt{2}} \sin x\right) \\ & = \sqrt{2} \left(\sin \dfrac{\pi}{4} \cos x + \cos \dfrac{\pi}{4} \sin x\right) \\ & = \sqrt{2} \sin \left(\dfrac{\pi}{4} + x\right) \end{aligned}$

$\therefore$ $\;$ $f'\left(x\right) = 0$ $\implies$ $\sin \left(\dfrac{\pi}{4} + x\right) = 0$

i.e. $\sin \left(\dfrac{\pi}{4} + x\right) = \sin 0 \;\; \text{OR} \;\; \sin \pi \;\; \text{OR} \;\; \sin 2 \pi$

$\implies$ $x = - \dfrac{\pi}{4}$ OR $x = \pi - \dfrac{\pi}{4} = \dfrac{3\pi}{4}$ OR $x = 2 \pi - \dfrac{\pi}{4} = \dfrac{7\pi}{4}$

Since $0 < x < 2 \pi$, $\;$ $x = - \dfrac{\pi}{4}$ is not a valid solution.

The points $x = \dfrac{3\pi}{4}$ and $x = \dfrac{7\pi}{4}$ divide the interval $0 < x < 2 \pi$ into three intervals namely $\left(0,\dfrac{3\pi}{4}\right)$, $\left(\dfrac{3\pi}{4},\dfrac{7\pi}{4}\right)$ and $\left(\dfrac{7\pi}{4},2\pi\right)$

Interval	Sign of $\sin \left(\dfrac{\pi}{4} + x\right)$	Nature of function f
$\left(0,\dfrac{3\pi}{4}\right)$	+ve	Increasing
$\left(\dfrac{3\pi}{4},\dfrac{7\pi}{4}\right)$	-ve	Decreasing
$\left(\dfrac{7\pi}{4}, 2 \pi\right)$	+ve	Increasing

$\therefore$ $f\left(x\right)$ is increasing in the interval $\left\{x : 0< x < \dfrac{3\pi}{4}\right\} \cup \left\{x : \dfrac{7\pi}{4}< x < 2\pi\right\} $

$f\left(x\right)$ is decreasing in the interval $\left\{x : \dfrac{7\pi}{4} < x < 2\pi\right\} $

Application of Derivatives: Increasing Decreasing Functions

Find the values of x for which $y = \left[x\left(x-2\right)\right]^2$ is an increasing function.

$y = \left[x\left(x-2\right)\right]^2$

$\therefore$ $\dfrac{dy}{dx}= 2 x \left(x-2\right) \left(x+x-2\right)$

i.e. $\dfrac{dy}{dx}= 4x \left(x-2\right) \left(x-1\right)$

$\dfrac{dy}{dx}=0$ $\implies$ $x=0$, $x=2$, $x=1$

These points divide the real line into four disjoint intervals namely $\left(-\infty,0\right)$, $\left(0,1\right)$, $\left(1,2\right)$ and $\left(2,\infty\right)$

Interval	Sign of $\dfrac{dy}{dx}$	Nature of function y
$\left(-\infty,0\right)$	-ve	Decreasing
$\left(0,1\right)$	+ve	Increasing
$\left(1,2\right)$	-ve	Decreasing
$\left(2,\infty\right)$	+ve	Increasing

$\therefore$ y is increasing in the intervals $\left(0,1\right)$ and $\left(2,\infty\right)$.

Application of Derivatives: Increasing Decreasing Functions

Find the intervals in which the function $f\left(x\right) = \left(x+1\right)^2 \left(x-3\right)^3$ is increasing or decreasing.

$f\left(x\right) = \left(x+1\right)^2 \left(x-3\right)^3$

$\therefore$ $f'\left(x\right) = 3 \left(x+1\right)^2 \left(x-3\right)^2 + 2 \left(x+1\right) \left(x-3\right)^3$

i.e. $f'\left(x\right) = \left(x+1\right) \left(x-3\right)^2 \left[3\left(x+1\right) +2 \left(x-3\right)\right]$

i.e. $f'\left(x\right) = \left(x+1\right) \left(x-3\right)^2 \left(5x-3\right)$

Now, $\left(x-3\right)^2 > 0 \; \forall \; x$

For critical points, $f'\left(x\right) = 0$

$\implies$ $x+1 = 0$ $\implies$ $x=-1$

or $\left(x-3\right)^2 =0$ $\implies$ $x=3$

or $5x-3=0$ $\implies$ $x=\dfrac{3}{5}$

These points divide the real line into four intervals namely $\left(-\infty, -1\right)$, $\left(-1, \dfrac{3}{5}\right)$, $\left(\dfrac{3}{5},3\right)$ and $\left(3,\infty\right)$

Interval	Sign of $\left(x+1\right)$	Sign of $\left(5x-3\right)$	Sign of $f'\left(x\right)$	Nature of function f
$\left(-\infty,-1\right)$	-ve	-ve	+ve	Increasing
$\left(-1,\dfrac{3}{5}\right)$	+ve	-ve	-ve	Decreasing
$\left(\dfrac{3}{5},3\right)$	+ve	+ve	+ve	Increasing
$\left(3,\infty\right)$	+ve	+ve	+ve	Increasing

$\therefore$ $f\left(x\right)$ is increasing in the interval $\left\{x : x<-1\right\} \cup \left\{x : x>\dfrac{3}{5}\right\} $

$f\left(x\right)$ is decreasing in the interval $\left\{x : -1 < x < \dfrac{3}{5}\right\} $

Application of Derivatives: Increasing Decreasing Functions

Find the least value of $a$ such that the function f given by $f\left(x\right) = x^2 +ax + 1$ is strictly increasing on $\left(1,2\right)$.

$f\left(x\right) = x^2 + ax + 1$

i.e. $f'\left(x\right) = 2x + a$

Since $f\left(x\right)$ is a strictly increasing function $\implies$ $f'\left(x\right) > 0$

i.e. $2 x + a > 0$ $\implies$ $x > -\dfrac{a}{2}$

i.e. To find $x > \dfrac{-a}{2}$ when x is in the interval $\left(1,2\right)$

i.e. To find $x > \dfrac{-a}{2}$ when $1 < x < 2$

Least value of x is 1.

$\therefore$ Least value of a is obtained when $x = 1$

$\therefore$ We have, $1 = \dfrac{-a}{2}$ $\implies$ $a = -2$

$\therefore$ The least value of $a = -2$.

Application of Derivatives: Increasing Decreasing Functions

Find the intervals in which the function $f:R \rightarrow R$, $f\left(x\right) = \dfrac{4 \sin x -2x - x \cos x}{2+\cos x}$ is increasing and in which it is decreasing.

$f\left(x\right) = \dfrac{4 \sin x -2x - x \cos x}{2 + \cos x}$

i.e. $f\left(x\right) = \dfrac{4\sin x}{2+\cos x} - \dfrac{x \left(2+\cos x\right)}{2+\cos x}$

i.e. $f\left(x\right) = \dfrac{4\sin x}{2+\cos x} - x$

$\therefore$ $f'\left(x\right) = \dfrac{\left(2+\cos x\right)\times 4 \cos x -4 \sin x \times \left(-\sin x\right)}{\left(2+\cos x\right)^2} -1$

i.e. $f'\left(x\right) = \dfrac{8 \cos x + 4 \cos ^2 x + 4 \sin ^2 x}{\left(2+\cos x\right)^2}-1$

i.e. $f'\left(x\right) = \dfrac{8 \cos x + 4 -4 - 4\cos x -\cos^2 x}{\left(2+\cos x\right)^2}$

i.e. $f'\left(x\right) = \dfrac{4 \cos x - \cos^2 x}{\left(2+\cos x\right)^2}$

i.e. $f'\left(x\right) = \dfrac{\cos x \left(4- \cos x\right)}{\left(2+\cos x\right)^2}$ $\;\; \cdots$ (1)

In equation (1), $\left(2+\cos x\right)^2 > 0 \; \forall \; x$ $\;\; \cdots$ (2)

Also, $4 - \cos x > 0 \; \forall \; x$ since $\cos x$ varies between $\pm 1$ $\;\; \cdots$ (3)

Case (i): $\mathbf{f\left(x\right)}$ is increasing:

$f\left(x\right)$ is increasing when $f'\left(x\right) > 0$

$\therefore$ From equation (1), for $f'\left(x\right) > 0$, $\cos x > 0$

Now, $\cos x > 0$ i.e. positive in the First and Fourth quadrants.

$\therefore$ $\cos x > 0$ $\implies$ $0 < x < \dfrac{\pi}{2}$ (First Quadrant) and $\dfrac{3\pi}{2} < x < 2\pi$ (Fourth Quadrant)

$\therefore$ Generalizing,

$\cos x > 0$ $\implies$ $2k\pi < x < \left(4k+1\right)\dfrac{\pi}{2}$ (First Quadrant) and $\left(4k+3\right) \dfrac{\pi}{2} < x < \left(2k+2\right)\pi$ (Fourth Quadrant), $k \in Z$

$\therefore$ $f\left(x\right)$ is increasing in the interval $\left(2k\pi, \left(4k+1\right)\dfrac{\pi}{2}\right)$ and $\left(\left(4k+3\right)\dfrac{\pi}{2}, \left(2k+2\right)\pi\right)$, $k \in Z$

Case (ii): $\mathbf{f\left(x\right)}$ is decreasing:

$f\left(x\right)$ is decreasing when $f'\left(x\right) < 0$

$\therefore$ In view of equations (2) and (3), we have from equation (1), for $f'\left(x\right) < 0$, $\cos x < 0$

Now, $\cos x < 0$ i.e. negative in the Second and Third quadrants.

$\therefore$ $\cos x < 0$ $\implies$ $\dfrac{\pi}{2} < x < \pi$ (Second Quadrant) and $\pi < x < \dfrac{3\pi}{2}$ (Third Quadrant)

$\therefore$ Generalizing,

$\cos x < 0$ $\implies$ $\left(4k+1\right)\dfrac{\pi}{2} < x < \left(2k+1\right)\pi$ (Second Quadrant) and $\left(2k+1\right)\pi < x < \left(4k+3\right) \dfrac{\pi}{2}$ (Third Quadrant), $k \in Z$

$\therefore$ $f\left(x\right)$ is decreasing in the interval $\left(\left(4k+1\right)\dfrac{\pi}{2}, \left(2k+1\right)\pi \right)$ and $\left(\left(2k+1\right) \pi, \left(4k+3\right)\dfrac{\pi}{2}\right)$, $k \in Z$

Application of Derivatives: Increasing Decreasing Functions

Discuss the nature of the function f given by $f\left(x\right) = \log \left[\sin \left(x\right)\right]$ in the intervals $\left(0,\dfrac{\pi}{2}\right)$ and $\left(\dfrac{\pi}{2}, \pi\right)$.

$f\left(x\right) = \log \left[\sin \left(x\right)\right]$

$\therefore$ $f'\left(x\right) = \dfrac{\cos x}{\sin x} = \cot x$

For $0 < x < \dfrac{\pi}{2}$, $\cot x > 0$

i.e. $f'\left(x\right) > 0$ for $0 < x < \dfrac{\pi}{2}$

i.e. $f\left(x\right)$ is strictly increasing in the interval $\left(0,\dfrac{\pi}{2}\right)$

For $\dfrac{\pi}{2} < x < \pi$, $\cot x < 0$

i.e. $f'\left(x\right) < 0$ for $\dfrac{\pi}{2} < x < \pi$

i.e. $f\left(x\right)$ is strictly decreasing in the interval $\left(\dfrac{\pi}{2}, \pi\right)$

Application of Derivatives: Increasing Decreasing Functions

Find the intervals in which $f\left(x\right) = \dfrac{3}{10}x^4-\dfrac{4}{5}x^3 -3x^2 + \dfrac{36}{5}x + 11$ is increasing and in which it is decreasing.

$f\left(x\right) = \dfrac{3}{10}x^4 - \dfrac{4}{5}x^3 - 3x^2 + \dfrac{36}{5}x + 11$

$\therefore$ $f'\left(x\right) = \dfrac{6}{5}x^3-\dfrac{12}{5}x^2-6x+\dfrac{36}{5}$

i.e. $f'\left(x\right) = \dfrac{6}{5}\left(x^3 -2x^2-5x+6\right)$

i.e. $f'\left(x\right) = \dfrac{6}{5}\left(x-1\right) \left(x^2 -x-6\right)$

i.e. $f'\left(x\right) = \dfrac{6}{5}\left(x-1\right) \left(x-3\right) \left(x+2\right)$

$\therefore$ For critical points, $f'\left(x\right) = 0$

$\implies$ $\left(x-1\right) \left(x-3\right) \left(x+2\right) = 0$

$\implies$ $x = -2$, $x=1$, $x=3$

These points divide the real line into four disjoint intervals namely $\left(-\infty,-2\right)$, $\left(-2,1\right)$, $\left(1,3\right)$ and $\left(3,\infty\right)$

Interval	Sign of $f'\left(x\right)$	Nature of function f
$\left(-\infty,-2\right)$	$< 0$	Decreasing
$\left(-2,1\right)$	$> 0$	Increasing
$\left(1,3\right)$	$ < 0$	Decreasing
$\left(3,\infty\right)$	$ > 0$	Increasing

$\therefore$ $f\left(x\right)$ is increasing in the intervals $\left(-2,1\right)$ and $\left(3,\infty\right)$

$f\left(x\right)$ is decreasing in the intervals $\left(-\infty,-2\right)$ and $\left(1,3\right)$

Application of Derivatives: Increasing Decreasing Functions

Prove that the function given by $f\left(x\right) = x^3 - 3x^2 +3x - 100$ is increasing in $\mathbb{R}$.

$f\left(x\right) = x^3 -3x^2 + 3x -100$

$\begin{aligned} \therefore f'\left(x\right) & = 3x^2 -6x +3 \\ & = 3 \left(x^2 -2x +1\right) \\ & = 3 \left(x-1\right)^2 \end{aligned}$

Now, $\left(x-1\right)^2 > 0 \; \forall \; \mathbb{R}$

$\therefore$ $3 \left(x-1\right)^2 > 0 \; \forall \; \mathbb{R}$

$\implies$ $f'\left(x\right) > 0 \; \forall \; \mathbb{R}$

$\implies$ $f\left(x\right)$ is strictly increasing in $\mathbb{R}$.

Application of Derivatives: Maxima and Minima

Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.

Let the two positive numbers be x and y.

Given: $x+y=16$

$\implies$ $y = 16-x$ $\;\; \cdots$ (1)

Sum of cubes of the numbers $= S = x^3 + y^3$

i.e. $S = x^3 + \left(16-x\right)^3$ [From equation (1)]

For minimum sum, $\dfrac{dS}{dx}=0$

Now, $\dfrac{dS}{dx}= 3x^2 - 3 \left(16-x\right)^2$ $\;\; \cdots$ (2)

$\therefore$ $\dfrac{dS}{dx}=0$ $\implies$ $3x^2 - 3 \left(16-x\right)^2 = 0$

i.e. $x^2 = \left(16-x\right)^2$

i.e. $x^2 = 256 -32x + x^2$

i.e. $32x = 256$ $\implies$ $x = 8$

From equation (2), $\dfrac{d^2 S}{dx^2} = 6x +6 \left(16-x\right) = 96 > 0 \; \forall \; x$

$\implies$ $x = 8$ gives a minimum for the sum of cubes of the numbers.

From equation (1), when $x = 8$, $y=16-8=8$

$\therefore$ The two numbers are 8, 8.

Application of Derivatives: Maxima and Minima

What fraction of the volume of a sphere is taken up by the largest cylinder that can be fit inside the sphere?

Let radius of sphere $=R$

Let radius of cylinder $=r$

Let height of cylinder $=2h$

From the figure, $R^2 = h^2 + r^2$

$\therefore$ $r^2 = R^2 - h^2$ $\;\; \cdots$ (1)

Volume of sphere $=V_s = \dfrac{4}{3}\pi R^3$ $\;\; \cdots$ (2)

Volume of cylinder $= V_c = 2\pi r^2 h$ $\;\; \cdots$ (3)

In view of equation (1), equation (3) becomes

$V_c = 2 \pi h \left(R^2 - h^2\right)$

i.e. $V_c = 2\pi R^2 h - 2 \pi h^3$ $\;\; \cdots$ (4)

For cylinder with maximum volume, $\dfrac{dV_c}{dh} = 0$

Now, from equation (4), $\dfrac{dV_c}{dh} = 2 \pi R^2 - 6\pi h^2$ $\;\; \cdots$ (5)

$\therefore$ $\dfrac{dV_c}{dh} = 0$ $\implies$ $2\pi R^2 - 6\pi h^2 = 0$

i.e. $3h^2 = R^2$ $\implies$ $h = \dfrac{R}{\sqrt{3}}$ $\;\; \cdots$ (6)

Now, from equation (5), $\dfrac{d^2 V_c}{dh^2} = -12 \pi h$

$\therefore$ $\dfrac{d^2 V_c}{dh^2} \bigg |_{h=\frac{R}{\sqrt{3}}} = \dfrac{-12\pi R}{\sqrt{3}} < 0$

$\implies$ $h = \dfrac{R}{\sqrt{3}}$ gives a cylinder with maximum volume.

Now, from equation (1), when $h = \dfrac{R}{\sqrt{3}}$,

$r = \sqrt{R^2 - \dfrac{R^2}{3}} = \dfrac{R\sqrt{2}}{\sqrt{3}}$ $\;\; \cdots$ (7)

$\therefore$ From equation (2), volume of cylinder $=V_c = 2 \pi \times \dfrac{2R^2}{3} \times \dfrac{R}{\sqrt{3}}$

i.e. $V_c = \dfrac{4\pi R^3}{3\sqrt{3}}$ $\;\; \cdots$ (8)

$\therefore$ From equations (2) and (8),

$\dfrac{V_c}{V_s} = \dfrac{4\pi R^3 / 3\sqrt{3}}{4\pi R^3 / 3} = \dfrac{1}{\sqrt{3}}$

$\therefore$ Fraction of the volume of a sphere is taken up by the largest cylinder that can be fit inside the sphere $= \dfrac{1}{\sqrt{3}} \times 100 = 57.5 \%$

Application of Derivatives: Maxima and Minima

A wire of length 36 cm is cut into two pieces. One of the pieces is turned in the form of a square and the other in the form of an equilateral triangle. Find the length of each piece so that the sum of the areas of the two be minimum.

Let the length of one piece of wire $= x$ cm

Then, length of remaining wire $= 36 - x$ cm

Let the wire of length x be turned in the form of a square.

$\therefore$ Length of side of square $= \dfrac{x}{4}$ cm

$\therefore$ Area of square $= \left(\dfrac{x}{4}\right)^2 = \dfrac{x^2}{16} \; cm^2$

$\left(36-x\right)$ cm wire is shaped as an equilateral triangle.

$\therefore$ Length of each side of triangle $=\dfrac{36-x}{3}$ cm

$\therefore$ Area of equilateral triangle $= \dfrac{\sqrt{3}}{4} \times \left(\dfrac{36-x}{3}\right)^2 = \dfrac{\sqrt{3}}{36} \left(36-x\right)^2 \; cm^2$

Sum of areas $= A = \dfrac{x^2}{16} + \dfrac{\sqrt{3}}{36} \left(36-x\right)^2$

For minimum area, $\dfrac{dA}{dx}=0$

Now, $\dfrac{dA}{dx} = \dfrac{2x}{16} + \dfrac{\sqrt{3}}{36} \times 2 \times \left(36-x\right) \times \left(-1\right)$

i.e. $\dfrac{dA}{dx} = \dfrac{x}{8} - \dfrac{\sqrt{3}}{18} \left(36-x\right)$ $\;\; \cdots$ (1)

$\therefore$ $\dfrac{dA}{dx} = 0$ $\implies$ $\dfrac{x}{8}- \dfrac{\sqrt{3}}{18} \left(36-x\right) = 0$

i.e. $9x = 4\sqrt{3}\left(36-x\right)$ $\implies$ $x = \dfrac{144 \sqrt{3}}{9+4\sqrt{3}}$

From equation (1), $\dfrac{d^2A}{dx^2} = \dfrac{1}{8} + \dfrac{\sqrt{3}}{18} > 0$

$\implies$ $x = \dfrac{144\sqrt{3}}{9+4\sqrt{3}}$ gives minimum area.

Length of second piece of wire $= 36 - \dfrac{144\sqrt{3}}{9+4\sqrt{3}} = \dfrac{324}{9+4\sqrt{3}}$

$\therefore$ Length of pieces of wire are $\dfrac{144\sqrt{3}}{9+4\sqrt{3}}$ cm and $\dfrac{324}{9+4\sqrt{3}}$ cm

Application of Derivatives: Maxima and Minima

Find the area of the largest rectangle that fits inside a semicircle of radius 10 units. One side of the rectangle is along the diameter of the semicircle.

Let radius of semicircle $= r$

Let length of rectangle $= 2\ell$

Let breadth of rectangle $= b$

From the figure, $r^2 = \ell^2 + b^2$

$\implies$ $\ell = \sqrt{r^2 - b^2}$ $\;\; \cdots$ (1)

Area of rectangle $=A= 2\ell b$

Substituting the value of $\ell$ from equation (1) gives

$A = 2b \sqrt{r^2 -b^2}$

For maximum area, $\dfrac{dA}{db}=0$

Now, $\dfrac{dA}{db} = 2\sqrt{r^2 - b^2} + \dfrac{2b}{2 \sqrt{r^2-b^2}}\times \left(-2b\right)$

i.e. $\dfrac{dA}{db} = 2 \sqrt{r^2 - b^2} - \dfrac{2b^2}{\sqrt{r^2 - b^2}}$

i.e. $\dfrac{dA}{db} = \dfrac{2r^2 - 4b^2}{\sqrt{r^2 - b^2}}$ $\;\; \cdots$ (2)

$\therefore$ $\dfrac{dA}{db} = 0$ $\implies$ $\dfrac{2r^2 -4b^2}{\sqrt{r^2 - b^2}} = 0$

i.e. $2b^2 = r^2$

$\implies$ $b = \dfrac{r}{\sqrt{2}}$

Substituting the value of b in equation (1) gives

$\ell = \sqrt{r^2 - \dfrac{r^2}{2}} = \dfrac{r}{\sqrt{2}}$

Now from equation (2) we have

$\dfrac{d^2 A}{db^2} = \dfrac{\sqrt{r^2 - b^2}\left(-2b\right)-\left(2r^2-4b^2\right) \times \dfrac{\left(-2b\right)}{2\sqrt{r^2-b^2}}}{r^2 - b^2}$

i.e. $\dfrac{d^2A}{db^2} = \dfrac{2b\left(r^2-2b^2-r^2+b^2\right)}{\left(r^2-b^2\right)\sqrt{r^2-b^2}}$

i.e. $\dfrac{d^2 A}{db^2} = \dfrac{-2b^3}{\left(r^2 - b^2\right) \sqrt{r^2 - b^2}}$

$\therefore$ $\dfrac{d^2A}{db^2} \bigg |_{b=\frac{r}{\sqrt{2}}} = \dfrac{\dfrac{-2r^3}{2\sqrt{2}}}{\left(r^2 - \dfrac{r^2}{2}\right)\sqrt{r^2 - \dfrac{r^2}{2}}} $ $= \dfrac{\dfrac{-2r^3}{2\sqrt{2}}}{\dfrac{r^3}{2\sqrt{2}}} = -2 < 0$

$\implies$ Area A is maximum when $\ell = \dfrac{r}{\sqrt{2}}$ and $b = \dfrac{r}{\sqrt{2}}$

$\therefore$ Maximum area of the rectangle $=2 \times \dfrac{r}{\sqrt{2}} \times \dfrac{r}{\sqrt{2}} = r^2$

Given: radius of semicircle $= r = 10$ units

$\therefore$ Maximum area of the rectangle $= 100$ sq units

Application of Derivatives: Maxima and Minima

Find the points at which the function f given by $f\left(x\right) = \left(x-2\right)^4 \left(x+1\right)^3$ has local maxima, local minima and point of inflection.

$f\left(x\right) = \left(x-2\right)^4 \left(x+1\right)^3$

$\begin{aligned} \therefore f'\left(x\right) & = 4 \left(x-2\right)^3 \left(x+1\right)^3 + 3 \left(x+1\right)^2 \left(x-2\right)^4 \\ & = \left(x-2\right)^3 \left(x+1\right)^2 \left[4\left(x+1\right)+3\left(x-2\right)\right] \\ & = \left(x-2\right)^3 \left(x+1\right)^2 \left(7x-2\right) \end{aligned}$

For critical points, $f'\left(x\right) = 0$

i.e. $\left(x-2\right)^3 \left(x+1\right)^2 \left(7x-2\right) = 0$

$\implies$ $x-2=0$ or $x+1=0$ or $7x-2=0$

i.e. $x=2$ or $x=-1$ or $x = \dfrac{2}{7}$

Consider $x=2$

When $x<2$, let $x=1.9$

$\begin{aligned} \text{Then, } f'\left(x\right) = f'\left(1.9\right) & = \left(1.9-2\right)^3 \left(1.9+1\right)^2 \left[\left(7\times 1.9\right) -2\right] \\ & = \left(-ve\right) \left(+ve\right) \left(+ve\right) \\ & = -ve \end{aligned}$

When $x>2$, let $x=2.1$

$\begin{aligned} \text{Then, } f'\left(x\right) = f'\left(2.1\right) & = \left(2.1-2\right)^3 \left(2.1+1\right)^2 \left[\left(7\times 2.1\right) -2\right] \\ & = \left(+ve\right) \left(+ve\right) \left(+ve\right) \\ & = +ve \end{aligned}$

i.e. $f'\left(x\right)$ changes sign from negative to positive as x increases through 2.

$\implies$ $x=2$ is a point of local minimum.

Consider $x=-1$

When $x<-1$, let $x=-1.1$

$\begin{aligned} \text{Then, } f'\left(x\right) = f'\left(-1.1\right) & = \left(-1.1-2\right)^3 \left(-1.1+1\right)^2 \left[7 \times\left(-1.1\right) -2\right] \\ & = \left(-ve\right) \left(+ve\right) \left(-ve\right) \\ & = +ve \end{aligned}$

When $x>-1$, let $x=-0.9$

$\begin{aligned} \text{Then, } f'\left(x\right) = f'\left(-0.9\right) & = \left(-0.9-2\right)^3 \left(-0.9+1\right)^2 \left[7 \times \left(-0.9\right) -2\right] \\ & = \left(-ve\right) \left(+ve\right) \left(-ve\right) \\ & = +ve \end{aligned}$

i.e. $f'\left(x\right)$ does not change sign as x increases through $-1$.

$\implies$ $x=-1$ is a point of inflexion.

Consider $x = \dfrac{2}{7}$

When $x < \dfrac{2}{7}$, let $x = \dfrac{1.9}{7}$

$\begin{aligned} \text{Then, } f'\left(x\right) = f'\left(\dfrac{1.9}{7}\right) & = \left(\dfrac{1.9}{7}-2\right)^3 \left(\dfrac{1.9}{7}+1\right)^2 \left[\left(7\times \dfrac{1.9}{7}\right) -2\right] \\ & = \left(-ve\right) \left(+ve\right) \left(-ve\right) \\ & = +ve \end{aligned}$

When $x>\dfrac{2}{7}$, let $x=\dfrac{2.1}{7}$

$\begin{aligned} \text{Then, } f'\left(x\right) = f'\left(\dfrac{2.1}{7}\right) & = \left(\dfrac{2.1}{7}-2\right)^3 \left(\dfrac{2.1}{7}+1\right)^2 \left[\left(7\times \dfrac{2.1}{7}\right) -2\right] \\ & = \left(-ve\right) \left(+ve\right) \left(+ve\right) \\ & = -ve \end{aligned}$

i.e. $f'\left(x\right)$ changes sign from positive to negative as x increases through $\dfrac{2}{7}$.

$\implies$ $x=\dfrac{2}{7}$ is a point of local maximum.

Application of Derivatives: Maxima and Minima

A window is in the form of a rectangle surmounted by semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window for maximum air flow through the window.

Let width of window $= \ell$

Then diameter of semicircle $= \ell$

Let height of window $= b$

Perimeter of the window $= P = \ell + 2b + \dfrac{\pi \ell}{2}$

Given: Perimeter of window $= 10 \; m$

$\implies$ $\ell + 2b + \dfrac{\pi \ell}{2} = 10$

i.e. $\ell \left(2+\pi\right) + 4b = 20$

i.e. $b = 5 - \dfrac{\ell \left(2+\pi\right)}{4}$ $\;\; \cdots$ (1)

Area of the window $=A= b \; \ell + \dfrac{\pi}{2} \times \dfrac{\ell^2}{4}$ $\;\; \cdots$ (2)

Substituting the value of b from equation (1) in equation (2) gives

$A = \ell \left[5- \dfrac{\ell \left(2+\pi\right)}{4}\right] + \dfrac{\pi \ell^2}{8}$

i.e. $A= 5 \ell - \dfrac{\pi}{4} \ell^2 -\dfrac{1}{2}\ell^2 + \dfrac{\pi}{8} \ell^2$

i.e. $A = 5 \ell - \dfrac{\pi}{8}\ell^2 - \dfrac{1}{2}\ell^2$

i.e. $A = 5 \ell - \dfrac{\ell^2}{2}\left(\dfrac{\pi}{4}+1\right)$

For maximum air flow, $\dfrac{dA}{d\ell} = 0$

i.e. $5-\ell\left(\dfrac{\pi+4}{4}\right) = 0$

$\implies$ $\ell = \dfrac{20}{\pi+4}$

Substituting the value of $\ell$ in equation (1) gives

$b = 5 - \left(\dfrac{\pi + 2}{4}\right) \left(\dfrac{20}{\pi + 4}\right)$

i.e. $b = 5 - 5 \left(\dfrac{\pi + 2}{\pi + 4}\right)$

i.e. $b = \dfrac{5\pi + 20 - 5\pi -10}{\pi + 4}$

i.e. $b = \dfrac{10}{\pi + 4}$

$\therefore$ Length of the window $\ell = \dfrac{20}{\pi + 4} \; m$; breadth of the window $=b=\dfrac{10}{\pi + 4} \; m$

Application of Derivatives: Maxima and Minima

The total cost of manufacturing x pocket radios per day is ₹ $\left(\dfrac{x^2}{4}+35x+25\right)$ and rate at which they may be sold to a distributor is ₹ $\left(\dfrac{100-x}{2}\right)$ each. What should be the daily output to attain a maximum total profit.

Daily output $=x$ radios

Cost price of x radios $=CP= ₹ \left(\dfrac{x^2}{4}+35x+25\right) $

Sale price of x radios $= SP= ₹ \; x \left(50 - \dfrac{x}{2}\right) = ₹ \left(50x - \dfrac{x^2}{2}\right)$

$\therefore$ Profit function $P\left(x\right) = SP - CP= \left(50x - \dfrac{x^2}{2} - \dfrac{x^2}{4} - 35x - 25\right)$

i.e. $P\left(x\right) = 15x - \dfrac{3x^2}{4}-25$

$\therefore$ For maximum profit, $\dfrac{dP}{dx} = 0$

Now, $\dfrac{dP}{dx} = 15 - \dfrac{6x}{4}$

$\therefore$ $\dfrac{dP}{dx}=0$ $\implies$ $15 - \dfrac{3}{2}x = 0$

i.e. $\dfrac{3}{2}x = 15$ $\implies$ $x = 10$

$\therefore$ Daily output $=10$ radios.

Application of Derivatives: Maxima and Minima

$f\left(x\right) = x^3 + 3ax^2 +3bx + c$ has a maximum at $x=-1$ and a minimum zero at $x=1$. Find a, b and c.

$f\left(x\right) = x^3 + 3ax^2 + 3bx +c$

$f'\left(x\right) = 3x^2 + 6ax + 3b$

For maxima or minima, $f'\left(x\right) = 0$

$\implies$ $3x^2 + 6ax+3b = 0$

i.e. $x^2 + 2ax + b = 0$ $\;\; \cdots$ (1)

Now, $f\left(x\right)$ has a minimum zero at $x=1$

$\implies$ $f\left(1\right) = 0$

i.e. $1+3a+3b+c = 0$

i.e. $3a+3b+c = -1$ $\;\; \cdots$ (2)

Further, $f\left(x\right)$ has a maximum at $x=-1$ and minimum at $x=1$

$\implies$ $x = \pm 1$ are roots of quadratic equation (1)

$\therefore$ From equation (1), sum of roots $=\dfrac{2a}{1} = 1-1$

i.e. $2a = 0$ $\implies$ $a=0$

Also, from equation (1), product of roots $= \dfrac{b}{1} = 1 \times \left(-1\right)$

i.e. $b = -1$

Substituting the values of a and b in equation (2) give

$-3+c=-1$ $\implies$ $c=2$

$\therefore$ $a=0$, $b=-1$, $c=2$

Application of Derivatives: Maxima and Minima

A container holding a fixed volume is made in the shape of a cylinder with a hemispherical top. The hemispherical top has the same radius as the cylinder. Find the ratio of height to radius of the cylinder which minimizes the cost of the container if

1. the cost per unit area of the top is twice as great as the cost per unit area of the side, and the container is made with no bottom;


2. the cost per unit area of the top is twice as great as the cost per unit area of the side, and the container is made with a circular bottom, for which the cost per unit area is 1.5 times the cost per unit area of the side.

Let radius of cylinder = radius of hemisphere = r

Let height of cylinder = h

Volume of container $= V = \pi r^2 h + \dfrac{2}{3}\pi r^3$

i.e. $V = \pi r^2 \left(h+\dfrac{2}{3}r\right)$

i.e. $h+\dfrac{2}{3}r = \dfrac{V}{\pi r^2}$

$\implies$ $h = \dfrac{V}{\pi r^2} - \dfrac{2}{3}r$ $\;\; \cdots$ (1)

1. Let cost per unit area of the side $= c$
   
   Then, cost per unit area of the hemispherical top $=2c$
   
   Area of the side of the container $=2\pi r h$
   
   Area of the top of the container $= 2 \pi r^2$
   
   $\therefore$ Cost of the container $= C = 2\pi r h c + 4 \pi r^2 c$ $\;\; \cdots$ (2)
   
   Substituting the value of h from equation (1) in equation (2) gives
   
   $C = 2 \pi r c \left(\dfrac{V}{\pi r^2} - \dfrac{2}{3}r\right) + 4 \pi r^2 c$
   
   i.e. $C = \dfrac{2cV}{r} - \dfrac{4}{3}\pi r^2 c + 4\pi r^2 c$
   
   i.e. $C = 2cV \times \dfrac{1}{r} + \dfrac{8\pi c}{3} \times r^2$ $\;\; \cdots$ (3)
   
   For minimum cost, $\dfrac{dC}{dr}=0$
   
   $\therefore$ From equation (3), $\dfrac{dC}{dr} = \dfrac{-2cV}{r^2}+\dfrac{16 \pi c r}{3} = 0$
   
   i.e. $\dfrac{16 \pi c r}{3} = \dfrac{2cV}{r^2}$
   
   i.e. $r^3 = \dfrac{3V}{8\pi}$
   
   $\implies$ $\dfrac{V}{\pi r^3} = \dfrac{8}{3}$ $\;\; \cdots$ (4)
   
   Now, from equation (1),
   
   $\dfrac{h}{r} = \dfrac{V}{\pi r^3}- \dfrac{2}{3}$ $\;\; \cdots$ (5)
   
   In view of equation (4), equation (5) can be written as
   
   $\dfrac{h}{r}= \dfrac{8}{3}-\dfrac{2}{3} = 2$
   
   $\therefore$ Ratio of height to radius of cylinder which minimizes the cost $= \dfrac{h}{r}=2:1$
   
   
2. Cost per unit area of the circular bottom of the container $= 1.5c$
   
   Area of the bottom of the container $= \pi r^2$
   
   $\therefore$ Cost of the container $=C = 4 \pi r^2 c + 2 \pi r h c + 1.5 \pi r^2 c$
   
   i.e. $C = 2\pi rhc + 5.5 \pi r^2 c$ $\;\; \cdots$ (6)
   
   Substituting the value of h from equation (1) in equation (6) gives
   
   $C= 2 \pi r c \left(\dfrac{V}{\pi r^2}-\dfrac{2}{3}r\right) + 5.5 \pi r^2 c$
   
   i.e. $C = \dfrac{2cV}{r} - \dfrac{4}{3}\pi r^2 c + \dfrac{11}{2}\pi r^2 c$
   
   i.e. $C = 2cV \times\dfrac{1}{r} + \dfrac{25\pi c}{6} \times r^2$ $\;\; \cdots$ (7)
   
   For minimum cost $\dfrac{dC}{dr} = 0$
   
   $\therefore$ We have from equation (7) for minimum cost
   
   $\dfrac{dC}{dr} = \dfrac{-2cV}{r^2} + \dfrac{50 \pi c r}{6} = 0$
   
   i.e. $\dfrac{2cV}{r^2} = \dfrac{25 \pi c r}{3}$
   
   $\implies$ $\dfrac{V}{\pi r^3} = \dfrac{25}{6}$ $\;\; \cdots$ (8)
   
   Substituting equation (8) in equation (5) gives
   
   $\dfrac{h}{r} = \dfrac{25}{6}- \dfrac{2}{3} = \dfrac{7}{2}$
   
   $\therefore$ Ratio of height to radius of cylinder which minimizes the cost $= \dfrac{h}{r}=7:2$