Question

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

• A. $\frac{\sin x+\cos x}{x\sin x+\cos x}+c$
• B. $\frac{x\sin x-\cos x}{x\sin x+\cos x}+c$
• C. $\frac{\sin x-x\cos x}{x\sin x+\cos x}+c$
• D. None of these

Answer 1

The correct option is C $\frac{\sin x-x\cos x}{x\sin x+\cos x}+c$

$I=\int \frac{x^{2}dx}{(x\sin x+\cos x{)}^2}$
$I=\int \left(\frac{x}{\cos x}\right)\left(\frac{x\cos x}{(x\sin x+\cos x{)}^2}\right)dx$
$=\left(\frac{x}{\cos x}\right)\int \left(\frac{x\cos x}{(x\sin x+\cos x{)}^2}\right)dx-\int \left[\frac{d}{dx}\left(\frac{x}{\cos x}\right)\int \left(\frac{x\cos x}{(x\sin x+\cos x{)}^2}\right)dx\right]dx...\left(1\right)$
Let $u=\int \left(\frac{x\cos x}{(x\sin x+\cos x{)}^2}\right)dx$
put $x\sin x+\cos x=t\Rightarrow x\cos xdx=dt$
$u=\int \frac{1}{t^2}dt=-\frac{1}{t}=-\frac{1}{x\sin x+\cos x}$
From equation $\left(1\right)$
$I=\left(\frac{x}{\cos x}\right)[-\frac{1}{x\sin x+\cos x}]-\int \left[\left(\frac{\cos x+x\sin x}{{\cos }^{2}x}\right)(-\frac{1}{x\sin x+\cos x})\right]dx$
$=-\frac{x}{\cos x(x\sin x+\cos x)}+\int {\sec }^{2}xdx$
$=-\frac{x}{\cos x(x\sin x+\cos x)}+\tan x+c$
$=\frac{\sin x-x\cos x}{x\sin x+\cos x}+c$