Question

$\int {\left(\frac{x}{x\sin x+\cos x}\right)}^{2}dx$ is equal to (where $C$ is a constant of integration)

• A. $\tan x-\frac{x\sec x}{x\sin x+\cos x}+C$
• B. $\sec x+\frac{x\tan x}{x\sin x+\cos x}+C$
• C. $\sec x-\frac{x\tan x}{x\sin x+\cos x}+C$
• D. $\tan x+\frac{x\sec x}{x\sin x+\cos x}+C$

Answer 1

The correct option is A $\tan x-\frac{x\sec x}{x\sin x+\cos x}+C$

$\int \frac{x^2}{(x\sin x+\cos x{)}^2}dx$
$\because \frac{d}{dx}(x\sin x+\cos x)=x\cos x$
using integration by parts method:
$=\int \frac{x\cos x}{(x\sin x+\cos x{)}^2}\left(\frac{x}{\cos x}\right)dx=\frac{x}{\cos x}\left[\frac{-1}{x\sin x+\cos x}\right]$

$-\int \frac{x\sin x+\cos x}{{\cos }^{2}x}\left[\frac{-1}{x\sin x+\cos x}\right]dx$
$=\frac{x}{\cos x}\left[\frac{-1}{x\sin x+\cos x}\right]+\int {\sec }^{2}xdx$
$=\frac{-x\sec x}{x\sin x+\cos x}+\tan x+C$