Q117 / 160 · AP EAMCET 2024 (22 May Shift 1) · Mathematics
$\int_{-\pi}^\pi \frac{x \sin x}{1+\cos ^2 x} d x=$
- A$\frac{3 \pi^2}{4}$
- B$\frac{\pi}{2}+1$
- C$\frac{\pi^2}{4}$
- Correct answerD$\frac{\pi^2}{2}$
Solution
$\begin{aligned}
& I=\int_{-\pi}^\pi \frac{x \sin x}{1+\cos ^2 x} d x ....(i)\\
& I=2 \int_0^\pi \frac{x \sin x}{1+\cos ^2 x} \\
& I=2 \int_0^\pi \frac{(\pi-x) \sin x}{1+\cos ^2 x} d x.....(ii)
\end{aligned}$
Adding (i) and (ii) $2 I=2 \pi \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x$
Let $\cos x=t \Rightarrow d t=-\sin x d x$ $I=\pi \int_{-1}^1 \frac{-1}{1+t^2} d t \Rightarrow I=2 \pi\left[\tan ^{-1} t\right]_0^1=\frac{\pi^2}{2}$
Adding (i) and (ii) $2 I=2 \pi \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x$
Let $\cos x=t \Rightarrow d t=-\sin x d x$ $I=\pi \int_{-1}^1 \frac{-1}{1+t^2} d t \Rightarrow I=2 \pi\left[\tan ^{-1} t\right]_0^1=\frac{\pi^2}{2}$
Hence, the correct answer is option: 'D'.