Question

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

Answer 1

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

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

Integrating by parts, taking $\frac{x}{\cos x}$ as the first function and $\frac{x\cos x}{{(x\sin x+\cos x)}^2}$ as the second function.

Now, let us first evaluate $\int \frac{x\cos xdx}{{(x\sin x+\cos x)}^2}$

Put $t=x\sin x+\cos x$

$\Rightarrow dt=(\sin x+x\cos x-\sin x)dx=x\cos xdx$ we get

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

$=\int \frac{dt}{t^2}$

$=\frac{-1}{t}=\frac{-1}{x\sin x+\cos x}$

Hence, $I=\frac{x}{\cos x}.\frac{-1}{x\sin x+\cos x}-\int \frac{\cos x+x\sin x}{{\cos }^{2}x}\times \frac{-1}{x\sin x+\cos x}dx$

$=\frac{x}{\cos x}.\frac{-1}{x\sin x+\cos x}-\int {\sec }^{2}xdx$

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

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

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

$=\frac{x(1-{\sin }^{2}x)+\sin x\cos x}{\cos x(x\sin x+\cos x)}+c$

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

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

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

$\therefore \int \frac{x^{2}dx}{{(x\sin x+\cos x)}^2}=\frac{(\sin x-x\cos x)}{(x\sin x+\cos x)}+c$