Question

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

Answer: (C)

$\frac{\sin x-x\cos x}{x\sin x+\cos x}+C$

Since, $\frac{d}{dx}(x\sin x+\cos x)=x\cos x+\sin x-\sin x=x\cos x$
$\therefore I=\int \frac{x^{2}dx}{(x\sin x+\cos x{)}^2}$
$=\int \frac{x}{\cos x}\cdot \frac{x\cos x}{(x\sin x+\cos x{)}^2}dx$

Solving by integration by parts,
$I=\frac{x}{\cos x}\cdot \left(\frac{-1}{x\sin x+\cos x}\right)-\int \frac{x\cos x+x\sin x}{{\cos }^{2}x}.\frac{-1}{(x\sin x+\cos x)}dx$
$=\frac{-x}{\cos x(x\sin x+\cos x)}+\int {\sec }^{2}xdx$
$=\frac{-x}{\cos x(x\sin x+\cos x)}+\tan x+C$

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

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