Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals5 Marks
Question
Evaluate : $\int_{-\pi}^\pi \frac{2 x(1+\sin x)}{1+\cos ^2 x}\ d x$
✓
Answer
Let
$I=\int_{-\pi}^\pi \frac{2 x(1+\sin x)}{1+\cos ^2 x} d x$
$I=\int_{-\pi}^\pi \frac{2 x+2 x \sin x}{1+\cos ^2 x} d x$
$I=\int_{-\pi}^\pi \frac{2 x d x}{1+\cos ^2 x}+\int_{-\pi}^\pi \frac{2 x \sin x}{1+\cos ^2 x} d x$
$I=I_1+I_2 \ldots \ldots(1)$
Here $I_1=\int_{-\pi}^\pi \frac{2 x d x}{1+\cos ^2 x}$ and $I _2$
$=\int_{-\pi}^\pi \frac{2 x \sin x}{1+\cos ^2 x} d x$
$\therefore \frac{2 x}{1+\cos ^2 x}$ is an odd function and $\frac{2 x \sin x}{1+\cos ^2 x}$ is an even function.
$\therefore I _1=0$ and $I _2=2 \int_0^\pi \frac{2 x \sin x}{1+\cos ^2 x} d x$
Now $ I _2=4 \int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x \ldots \ldots(2)$
Using property $P _5$ in $I _2$,
$\Rightarrow I _2=4 \int_0^\pi \frac{(\pi-x) \sin (\pi-x)}{1+\cos ^2(\pi-x)} d x$
$\left(\because \int_0^a f(x) d x=\int_0^a f(a-x) d x\right)$
$I_2=4 \int_0^\pi \frac{(\pi-x) \sin x}{1+\cos ^2 x} d x$
$I_2=4 \int_0^\pi \frac{\pi \sin x}{1+\cos ^2 x} d x-4 \int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x \ldots$
$I_2=4 \pi \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x-I_2$
$2 I_2=4 \pi \int_0^\pi \frac{\sin x d x}{1+\cos ^2 x}$
putting
$\cos x=t $
$\Rightarrow-\sin x \ d x=d t$
$\sin x\ d x=-d t$
$0 \ n$ changing limits when $x=0$, then $t=1$ and when $x=$ then $t=-1$
$2 I_2 =4 \pi \int_1^{-1} \frac{-d t}{1+t^2}$
$2 I_2 =4 \pi \int_{-1}^1 \frac{d t}{1+t^2}$
$I_2 =2 \pi\left(\tan ^{-1} t\right)_{-1}^1$
$ =2 \pi\left(\tan ^{-1} 1-\tan ^{-1}(-1)\right)$
$ =2 \pi\left(\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)\right)$
$($using property $P_2)$
$=2 \pi\left(\frac{\pi}{4}+\frac{\pi}{4}\right)$
$=2 \pi \times \frac{2 \pi}{4}=\frac{4 \pi^2}{4}$
$I_2 =\pi^2$
Putting $I _1$ and $I _2$ in eq. $(1)$
$I=0+\pi^2=\pi^2$
$\Rightarrow \int_{-\pi}^\pi \frac{2 x(1+\sin x)}{1+\cos ^2 x} d x=\pi^2$
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