Indefinite Integration 1 Question 72
73. Determine the value of $\int _{-\pi}^{\pi} \frac{2 x(1+\sin x)}{1+\cos ^{2} x} d x$.
$(1995,5 M)$
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Solution:
- Let } \quad I=\int _{-\pi}^{\pi} \frac{2 x(1+\sin x)}{1+\cos ^{2} x} d x \\ & \quad I=\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 \\ & \quad I=I _1+I _2 \end{aligned} $$
Now, $I _1=\int _{-\pi}^{\pi} \frac{2 x}{1+\cos ^{2} x} d x$
Let $f(x)=\frac{2 x}{1+\cos ^{2} x}$
$\Rightarrow f(-x)=\frac{-2 x}{1+\cos ^{2}(-x)}=\frac{-2 x}{1+\cos ^{2} x}=-f(x)$
$\Rightarrow \quad f(-x)=-f(x)$ which shows that $f(x)$ is an odd function.
$$ \begin{aligned} \therefore & I _1 & =0 \\ \text { Again, let } & g(x) & =\frac{2 x \sin x}{1+\cos ^{2} x} \end{aligned} $$
$\Rightarrow g(-x)=\frac{2(-x) \sin (-x)}{1+\cos ^{2}(-x)}=\frac{2 x \sin x}{1+\cos ^{2} x}=g(x)$
$\Rightarrow g(-x)=g(x)$ which shows that $g(x)$ is an even function.
$\therefore \quad I _2=\int _{-\pi}^{\pi} \frac{2 x \sin x}{1+\cos ^{2} x} d x=2 \cdot 2 \int _0^{\pi} \frac{x \sin x}{1+\cos ^{2} x} d x$
$$ \begin{aligned} & =4 \int _0^{\pi} \frac{(\pi-x) \sin (\pi-x)}{1+[\cos (\pi-x)]^{2}} d x=4 \int _0^{\pi} \frac{(\pi-x) \sin x}{1+\cos ^{2} x} d x \\ & =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 \end{aligned} $$
$\Rightarrow I _2=4 \pi \int _0^{\pi} \frac{\sin x}{1+\cos ^{2} x} d x-I _2$
$\Rightarrow 2 I _2=4 \pi \int _0^{\pi} \frac{\sin x}{1+\cos ^{2} x} d x$
Put $\quad \cos x=t \quad \Rightarrow \quad-\sin x d x=d t$
$\therefore \quad I _2=-2 \pi \int _1^{-1} \frac{d t}{1+t^{2}}=2 \pi \int _{-1}^{1} \frac{d t}{1+t^{2}}=4 \pi \int _0^{1} \frac{d t}{1+t^{2}}$
$$ =4 \pi\left[\tan ^{-1} t\right] _0^{1}=4 \pi\left[\tan ^{-1} 1-\tan ^{-1} 0\right] $$
$$ =4 \pi(\pi / 4-0)=\pi^{2} $$
$\therefore \quad I=I _1+I _2=0+\pi^{2}=\pi^{2}$
