Question

$\int \frac{x^2}{{(xsinx+cosx)}^2}$

Answer 1

$\int \frac{x^2}{{(x\sin x+\cos x)}^2}dx$

$=\int \left(x\sec x\right)\frac{x\cos x}{{(x\sin x+\cos x)}^2}dx$

$=x\sec x.\frac{-1}{x\sin x+\cos x}-\int (\sec x+x\sec x\tan x)\times \left(\frac{-1}{x\sin x+\cos x}\right)dx$

$=\left(x\sec x\right)\left(\frac{-1}{x\sin x+\cos x}\right)-\int (\sec x+x\sec x\tan x)\times \left(\frac{-1}{x\sin x+\cos x}\right)dx$

$=\frac{-x\sec x}{x\sin x+\cos x}+\int {\sec }^{2}xdx$

$=\frac{-x\sec x}{x\sin x+\cos x}+\tan x+c$

$=\frac{-1}{\cos x[x\sin x+\cos x]}+\tan x+c$

$=\frac{-x+{\sin }^{2}x(\sin x+\cos x)}{\cos x[x\sin x+\cos x]}+c$

$=\frac{\sin x-x\cos x}{x\sin x+\cos x}+c$