Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration2 Marks
MCQ
$\int_{-\pi}^\pi \frac{2 x(1+\sin x)}{1+\cos ^2 x} d x$ is
A
$\frac{\pi^2}{4}$
✓
$\pi^2$
C
$0$
D
$\frac{\pi}{2}$
✓
Answer
Correct option: B.
$\pi^2$
(B) Let $I =\int_{-\pi}^\pi \frac{2 x(1+\sin x)}{1+\cos ^2 x} d x$ $=\int_{-\pi}^\pi \frac{2 x}{1+\cos ^2 x} d x+\int_{-\pi}^\pi \frac{2 x \sin x}{1+\cos ^2 x} d x$ Since $\frac{2 x}{1+\cos ^2 x}$ is an odd function and $\frac{2 x \sin x}{1+\cos ^2 x}$ is an even function. $\therefore \quad I =0+2 \int_0^\pi \frac{2 x \sin x}{1+\cos ^2 x} d x$ $\Rightarrow I =4 \int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x$ ...(i) $\Rightarrow I =4 \int_{-\pi}^\pi \frac{(\pi-x) \sin x}{1+\cos ^2 x} d x$ ...(ii) $\ldots\left[\because \int_0^{ a } f (x) d x=\int_0^{ a } f ( a -x) d x\right]$ Adding (i) and (ii), we get $2 I =4 \int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x \Rightarrow I =2 \pi \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x$ Put $\cos x= t \Rightarrow-\sin x d x= dt$ $\therefore I=2 \pi \int_1^{-1} \frac{- dt }{1+ t ^2}$ $\Rightarrow I =-2 \pi\left[\tan ^{-1} t \right]_1^{-1}=-2 \pi\left(\frac{-\pi}{4}-\frac{\pi}{4}\right)=\pi^2$
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