If $I=\int \frac{x^2 d x}{(x \sin x+\cos x)^2}=f(x)+\tan x+c$, then $f(x)$ is

Question

If $I=\int \frac{x^2 d x}{(x \sin x+\cos x)^2}=f(x)+\tan x+c$, then $f(x)$ is, $\frac{\sin x}{x \sin x+\cos x}$, $\frac{1}{(x \sin x+\cos x)^2}$, $\frac{-x}{\cos x(x \sin x+\cos x)}$, $\frac{1}{\sin x(x \cos x+\sin x)}$

MARKS App · Accepted Answer

Hint : $I=\int \frac{x^2}{(x \sin x+\cos x)^2} d x=\int \frac{x}{(x \sin x+\cos x)^2} \times \frac{x}{\cos x} d x$ $I=-\frac{1}{(x \sin x+\cos x)} \cdot \frac{x}{\cos x}+\int \frac{1}{(x \sin x+\cos x)} \cdot \frac{(\cos x+x \sin x)}{\cos ^2 x} d x$ $\begin{aligned} & I=-\frac{1}{(x \sin x+\cos x)} \cdot \frac{x}{\cos x}+\int \sec ^2 x d x \ & I=-\frac{x}{(x \sin x+\cos x) \cos x}+\tan x+C \ & \therefore f(x)=\frac{-x}{\cos x(x \sin x+C)}\end{aligned}$

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