$\int_{-\pi}^\pi \frac{x \sin x}{1+\cos ^2 x} d x=$

$\int_{-\pi}^\pi \frac{x \sin x}{1+\cos ^2 x} d x=$
  1. $\frac{3 \pi^2}{4}$
  2. $\frac{\pi}{2}+1$
  3. $\frac{\pi^2}{4}$
  4. $\frac{\pi^2}{2}$

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 $\cos x=t \Rightarrow d t=-\sin x d x$ $I=\pi \int_{-1}^1 \frac{-1}{1+t^2} d t \Rightarrow I=2 \pi\left[\tan ^{-1} t\right]_0^1=\frac{\pi^2}{2}$

Asked in: AP EAMCET 2024 (22 May Shift 1)

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