Evaluate : _ - ^ 2 x(1+ x) 1+ ^2 x d x

Question

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

✓

Answer

Let
$
\begin{array}{l}
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
\end{array}
$
$
\begin{array}{l}
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\quad \quad \ldots \ldots(1)
\end{array}
$
Here $_1=\int_{-\pi}^\pi \frac{2 x d x}{1+\cos ^2 x}$ 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.
$\therefore \quad I _1=0$ and $I _2=2 \int_0^\pi \frac{2 x \sin x}{1+\cos ^2 x} d x$
Now $\quad I _2=4 \int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x\quad \quad \ldots \ldots(2)$
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)
$
$
\begin{array}{l}
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} \\
\text { putting } \begin{aligned}
\cos x=t \Rightarrow-\sin x d x=d t \\
\sin x d x=-d t
\end{aligned}
\end{array}
$
0 n changing limits when $x=0$, then $t=1$ and when $x=$ then $t=-1$
$
\begin{aligned}
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)
\end{aligned}
$ (using property $P _2$ )
$
\begin{aligned}
& =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
\end{aligned}
$
Putting $I _1$ and $I _2$ in eq. (1)
$
I=0+\pi^2=\pi^2
$
$\Rightarrow \int_{-\pi}^\pi \frac{2 x(1+\sin x)}{1+\cos ^2 x} d x=\pi^2$