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# Evaluate:$\large\int\limits_{-\pi}^\pi \frac {2x(1+\sin x)}{1+\cos ^2x}$

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A)
Toolbox:
• (i)$\int \limits _a^b f(x)dx=F(b)-F(a)$
• (ii) If $f(-x)=-f(x),$ then it is an odd function, then $\int \limits^a_{-a} f(x) dx=0$
• (iii)If $f(-x)=f(x),$ thenit is an even function, then $\int \limits_{-a}^a f(x)dx= 2 \int\limits _0^a f(x)dx$
• (iv)$\large\int \frac{dx}{x^2+a^2}=\frac{1}{a}$$\tan ^{-1}(x/a)+c Step 1: GivenI=\int\limits_{-\pi}^\pi \large\frac {2x(1+\sin x)}{1+\cos ^2x}$$dx=\int\limits_{-\pi}^\pi \large\frac {2x2\sin x}{1+\cos ^2x}$$dx on seperating the terms, \int\limits_{-\pi}^\pi \large\frac {2x}{1+\cos ^2x}$$dx+\int\limits_{-\pi}^\pi \frac {2x\sin x}{1+\cos ^2x}$$dx But \large\frac{2x}{1+\cos ^2x} is an odd function, because if x is replaced by -x we get \large\frac{-2x}{1+\cos ^2x} Hence \int\limits_{-\pi}^\pi \large\frac {2x}{1+\cos ^2x}$$dx=0\;Hence\;I_1=0$
Consider $\int\limits_{-\pi}^\pi \large\frac {2x\sin x}{1+\cos ^2x}$$dx is an even function because if x is replaced by -x we get \large\frac{2(-x)(\sin(-x))}{1+\cos^2(-x)}=\frac{2x\sin x}{1+\cos ^2 x}\qquad$$(\sin (-x)=-\sin x)$
Step 2:
Let $I_2=\large\int\limits_{-\pi}^\pi \frac {2x\sin x}{1+\cos ^2x}$$dx Since it is an even function we can write this as 2\int\limits_0^\pi \large\frac {2x\sin x}{1+\cos ^2x}$$dx$
$I_2=4\int\limits_0^\pi \large\frac {x\sin x}{1+\cos ^2x}$$dx-----(1) By applying the property \int\limits_a^a f(x)dx=\int \limits_0^a f(a-x)dx I_2=4\int\limits_0^\pi \large\frac {(\pi-x)\sin (\pi-x)}{1+\cos ^2(\pi-x)}$$dx$-----(2)
But $\sin (\pi-x)=\sin x$
$1+\cos ^2(\pi-x)=1+\cos ^2 x$
Adding equ(1) and (2) we get
$2I_2=4\int\limits_0^\pi \bigg[\large\frac {x\sin x}{1+\cos ^2x}+\frac {(\pi-x)\sin x}{1+\cos ^2x}\bigg]dx$
$=4\pi\int\limits_0^\pi \bigg[\large\frac {\sin x}{1+\cos ^2x}$$dx\bigg]_0^\pi Step 3: Let \cos x=t on differentiating w.r.t. x we get -\sin xdx=dt\qquad =>\sin x dx =-dt on substituting t and dt, we also know that limit changes when we substitute for t, when x=0,t=\cos \;0=1 when x=1,t=\cos\; \pi=-1 Hence 2I_2=4\pi\int\limits_1^{-1}\large\frac {-dt}{1+t^2} The negative symbol can be removed by changing the limits, Therefore 2I_2=4\pi\int\limits_{-1}^1 \large\frac {dt}{1+t^2} This is of the form \int \large\frac{dx}{x^2+a^2}=\frac{1}{a}$$\tan ^{-1}(\frac{x}{a})+c$
Therefore on integrating we get
$2I=4\pi[\tan ^{-1}(t)]_{-1}^1$
Step 4:
On applying limits we get,
$2I_2=4\pi[\tan^{-1}(1)-\tan^{-1}(-1)]$
But $\tan ^{-1}(1)=\pi/4 \; and\;\tan ^{-1}(-1)=-\pi/4$
Therefore $2I_2=4\pi[\large\frac{\pi}{4}-(\frac{-\pi}{4})]$
$2I_2=4\pi[\large\frac{\pi}{4}+\frac{\pi}{4}]$
$2I_2=4\pi(\large\frac{\pi}{2})$$=2 \pi ^2$
Therefore $I_2=\pi^2$
Hence $I=I_1+I_2$
$=0+\pi^2$
$I=\pi^2$