Question

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

Answer 1

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

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

Using the rule of integration by parts, we get

$\frac{d}{dx}\left(\frac{1}{x\sin x+\cos x}\right)$

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

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

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

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

$\int \frac{x\cos x}{{(x\sin x+\cos x)}^2}dx=\frac{-1}{x\sin x+\cos x}$ ..........$\left(1\right)$

Also,$\frac{d}{dx}\left(x\sec x\right)=x\sec x\tan x+\sec x$

$=x\times \frac{1}{\cos x}\times \frac{\sin x}{\cos x}+\frac{1}{\cos x}$

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

Now, in integration by parts, we take,

$u=x\sec x$ and $dv=\frac{x\cos x}{{(x\sin x+\cos x)}^2}$

$\therefore du=\frac{x\sin x+\cos x}{{\cos }^{2}x}$ and $v=\frac{-1}{x\sin x+\cos x}$

$\therefore I=\left(x\sec x\right)\left(\frac{-1}{x\sin x+\cos x}\right)-\int \frac{x\sin x+\cos x}{{\cos }^{2}x}\times \frac{-1}{x\sin x+\cos x}dx$

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

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

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

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

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

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

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

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

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

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