Q117 / 160 · AP EAMCET 2024 (20 May Shift 2) · Mathematics
$\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x=$
- A0
- B$\frac{\pi}{2}$
- C$\frac{\pi^2}{2}$
- Correct answerD$\frac{\pi^2}{4}$
Solution
$\mathrm{I}=\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x$ $\qquad ....\mathrm{(i)}$
$I=\int_0^\pi \frac{(\pi-x) \sin (\pi-x)}{1+\cos ^2(\pi-x)} d x=\int_0^\pi \frac{(\pi-x) \sin x}{\left.1+\cos ^2 x\right)} d x$ $\qquad ....\mathrm{(ii)}$
Adding (i) and (ii), $2 \mathrm{I}=\int_0^\pi \frac{\pi \sin x}{1+\cos ^2 x} d x$
Let $\cos x=t \Rightarrow-\sin x d x=d t$ $\begin{aligned} & 2 \mathrm{I}=\pi \int_{-1}^1 \frac{d t}{1+t^2} \\ & \Rightarrow 2 \mathrm{I}=\pi\left[\tan ^{-1}(t)\right]_{-1}^1 \Rightarrow \mathrm{I}=\frac{\pi^2}{4} \end{aligned}$
Adding (i) and (ii), $2 \mathrm{I}=\int_0^\pi \frac{\pi \sin x}{1+\cos ^2 x} d x$
Let $\cos x=t \Rightarrow-\sin x d x=d t$ $\begin{aligned} & 2 \mathrm{I}=\pi \int_{-1}^1 \frac{d t}{1+t^2} \\ & \Rightarrow 2 \mathrm{I}=\pi\left[\tan ^{-1}(t)\right]_{-1}^1 \Rightarrow \mathrm{I}=\frac{\pi^2}{4} \end{aligned}$
Hence, the correct answer is option: 'D'.