Question

If $\int \frac{dx}{\sqrt{{\sin }^{3}x{\cos }^{5}x}}=a\sqrt{\cot x}+b\sqrt{{\tan }^{3}x}+c,$ then

Answer 1

$\begin{array}{c}GivenI={\int }^{}\frac{dx}{\sqrt{{\sin }^{3}x{\cos }^{5}x}}\\ ={\int }^{}\frac{dx}{\sqrt{\frac{{\sin }^{3}x}{{\cos }^{3}x}{\cos }^{3}x}}\\ ={\int }^{}\frac{{\sec }^{4}x}{{\tan }^{\frac{3}{2}}x}dx\\ ={\int }^{}\frac{(1+{\tan }^{2}x){\sec }^{2}x}{{\tan }^{\frac{3}{2}}x}dx\\ Let\tan x=t\\ {\sec }^{2}xdx=dt\\ Hence,I={\int }^{}\frac{(1+t^2)}{t^{\frac{3}{2}}}dt\\ ={\int }^{}{t}^{\frac{-3}{2}}dt+{\int }^{}{t}^{\frac{1}{2}}dt\\ =-2\frac{1}{t^{\frac{1}{2}}}+\frac{2}{3}{t}^{\frac{3}{2}}+C\\ =-2\sqrt{\cot x}+\frac{2}{3}\sqrt{{\tan }^{3}x+}C\end{array}$