Question:
The integral $\int\left(\frac{x}{x \sin x+\cos x}\right)^{2} d x$ is equal to :
(where $\mathrm{C}$ is a constant of integration)
Correct Option: , 4
Solution:
$\int\left(\frac{x}{x \sin x+\cos x}\right)^{2} d x=\int\left(\frac{x}{\cos x}\right) \cdot \frac{x \cos x d x}{(x \sin x+\cos x)^{2}}$
$=\frac{x}{\cos x}\left(-\frac{1}{x \sin x+\cos x}\right)$
$+\int\left(\frac{\cos x+x \sin x}{\cos ^{2} x}\right)\left(\frac{1}{x \sin x+\cos x}\right) d x$
$=-\frac{x \sec x}{x \sin x+\operatorname{cox}}+\int \sec ^{2} x d x$
$=-\frac{x \sec x}{x \sin x+\operatorname{cox}}+\tan x+C$