Question

Solve : $\frac{dx}{{\sin }^{3}x.{\cos }^{5}x}$

Answer 1

$I=\int \frac{dx}{({sin}^{3}x.{cos}^{5}x)}$ substituting,\\ $\sin x=\frac{\tan x}{\sec x}\cos x\frac{1}{\sec x}{\sec }^{2}x={\tan }^{2}x+1$ $\therefore I=\int {\sec }^{2}x.\frac{{({\tan }^{2}x+1)}^3}{{\tan }^{3}x}dxLet,u=\tan x$ $\to dx=\frac{1}{{\sec }^2}du$ $\therefore I=\int \frac{{(u^2+1)}^3}{u^3}du=\int (u^3+3u+\frac{3}{u}+\frac{1}{u^3})du$ \\ $=\int u^{3}du+3\int udu+3\int \frac{1}{u}du+\int \frac{1}{u^3}du$ \\ $=3\log u+\frac{u^4}{4}+\frac{{3u}^2}{2}-\frac{1}{{2u}^2}$

=$3\log tanx$$+\frac{{\tan }^{4}x}{4}+\frac{3{\tan }^{2}x}{2}-\frac{1}{2{tan}^{2}x}+C$