WBJEE · Maths · Indefinite Integration
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 \(\frac{1}{(x \sin x+\cos x)^2}\)
- C \(\frac{-x}{\cos x(x \sin x+\cos x)}\)
- D \(\frac{1}{\sin x(x \cos x+\sin x)}\)
Answer & Solution
Correct Answer
(C) \(\frac{-x}{\cos x(x \sin x+\cos x)}\)
Step-by-step Solution
Detailed explanation
Hint : \(I=\int \frac{x^2}{(x \sin x+\cos x)^2} d x=\int \frac{x}{(x \sin x+\cos x)^2} \times \frac{x}{\cos x} d x\) \(I=-\frac{1}{(x \sin x+\cos x)} \cdot \frac{x}{\cos x}+\int \frac{1}{(x \sin x+\cos x)} \cdot \frac{(\cos x+x \sin x)}{\cos ^2 x} d x\)…
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