AP EAMCET · Maths · Definite Integration
\(\int_0^{\pi / 4} \frac{x^2}{(x \sin x+\cos x)^2} d x=\)
- A \(\frac{2-\pi}{2+\pi}\)
- B \(\frac{4-\pi}{4+\pi}\)
- C \(\frac{6-\pi}{6+\pi}\)
- D \(\frac{8-\pi}{8+\pi}\)
Answer & Solution
Correct Answer
(B) \(\frac{4-\pi}{4+\pi}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { } \quad I=\int \frac{x^2}{(x \sin x+\cos x)^2} d x \\ & =\int(x \sec x)\left(\frac{x \cos x}{(x \sin x+\cos x)^2}\right) d x \\ & =x \sec x \times \frac{(-1)}{(x \sin x+\cos x)} \\ & \quad-\int(\sec x+x \sec x \tan x) \frac{-1}{(x \sin x+\cos x)} d…
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