Q46 / 155 · WBJEE 2023 · Mathematics
If \(I=\int \frac{x^2 d x}{(x \sin x+\cos x)^2}=f(x)+\tan x+c\), then \(f(x)\) is
- A\(\frac{\sin x}{x \sin x+\cos x}\)
- B\(\frac{1}{(x \sin x+\cos x)^2}\)
- Correct answerC\(\frac{-x}{\cos x(x \sin x+\cos x)}\)
- D\(\frac{1}{\sin x(x \cos x+\sin x)}\)
Solution
Hint : \(I=\int \frac{x^2}{(x \sin x+\cos x)^2} d x=\int \frac{x}{(x \sin x+\cos x)^2} \times \frac{x}{\cos x} d x\)
\(I=-\frac{1}{(x \sin x+\cos x)} \cdot \frac{x}{\cos x}+\int \frac{1}{(x \sin x+\cos x)} \cdot \frac{(\cos x+x \sin x)}{\cos ^2 x} d x\)
\(\begin{aligned} & I=-\frac{1}{(x \sin x+\cos x)} \cdot \frac{x}{\cos x}+\int \sec ^2 x d x \\ & I=-\frac{x}{(x \sin x+\cos x) \cos x}+\tan x+C \\ & \therefore f(x)=\frac{-x}{\cos x(x \sin x+C)}\end{aligned}\)
Hence, the correct answer is option: 'C'.