Q96 / 225 · JEE Main 2002 · Mathematics

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

  • A
    $\frac{\pi^2}{4}$
  • Correct answer
    B
    $\pi^2$
  • C
    zero
  • D
    $\frac{\pi}{2}$

Solution

$$ \begin{aligned} & \int_{-\pi}^\pi \frac{2 x(1+\sin x)}{1+\cos ^2 x} d x=\int_{-\pi}^\pi \frac{2 x}{1+\cos ^2 x}+2 \int_{-\pi}^\pi \frac{x \sin x}{1+\cos ^2 x} \\ & =0+4 \int_0^\pi \frac{x \sin x d x}{1+\cos ^2 x} I=4 \int_0^\pi \frac{(\pi-x) \sin (\pi-x)}{1+\cos ^2(\pi-x)} \\ & I=4 \int_0^\pi \frac{(\pi-x) \sin x}{1+\cos ^2 x} \Rightarrow I=4 \pi \int_0^\pi \frac{\sin x}{1+\cos ^2 x}-4 \pi \int \frac{x \sin x}{1+\cos ^2 x} \Rightarrow 2 I=4 \pi \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x \end{aligned} $$ put $\cos x=t$ and solve it.

Hence, the correct answer is option: 'B'.

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