AP EAMCET · Maths · Definite Integration
\(\int_{-\pi}^\pi \frac{x \sin x}{1+\cos ^2 x} d x=\)
- A \(\frac{3 \pi^2}{4}\)
- B \(\frac{\pi}{2}+1\)
- C \(\frac{\pi^2}{4}\)
- D \(\frac{\pi^2}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{\pi^2}{2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & I=\int_{-\pi}^\pi \frac{x \sin x}{1+\cos ^2 x} d x ....(i)\\ & I=2 \int_0^\pi \frac{x \sin x}{1+\cos ^2 x} \\ & I=2 \int_0^\pi \frac{(\pi-x) \sin x}{1+\cos ^2 x} d x.....(ii) \end{aligned}\) Adding (i) and (ii)…
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