The integral


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)

  1. $\sec x+\frac{x \tan x}{x \sin x+\cos x}+C$

  2. $\sec x-\frac{x \tan x}{x \sin x+\cos x}+C$

  3. $\tan x+\frac{x \sec x}{x \sin x+\cos x}+C$

  4. $\tan x-\frac{x \sec x}{x \sin x+\cos x}+C$

Correct Option: , 4


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

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

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

$=-\frac{x \sec x}{x \sin x+\operatorname{cox}}+\int \sec ^{2} x d x$

$=-\frac{x \sec x}{x \sin x+\operatorname{cox}}+\tan x+C$


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