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