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Evaluate:$\int\limits _0 ^{\pi} \large\frac {x \sin x}{1+\cos ^2 x}$$dx$

1 Answer

  • $ \int \limits_a^b f(x)dx =F(b)-F(a)$
  • $\int \large\frac{dx}{x^2+a^2}=\frac{1}{a} $$\tan ^{-1} (x/a)$
  • If $f(x)$ is substituted by t, then $f'(x)dx=dt, $hence $\int f(x) dx=\int t.dt$
Step 1:
Given $I=\int \limits_0^{\pi} \large\frac{x \sin x}{1+\cos ^2 x} $$dx$
By applying the property, $\int \limits_0^a f(x)dx =\int \limits_0^a f(a-x) dx$
$I= \int \limits_0 ^\pi \large\frac{(\pi-x) \sin (\pi-x)}{1+\cos ^2 (\pi-x)}$$dx; $ we know $\sin (\pi-x)=\sin x\; \qquad\cos (\pi-x)=-\cos x$
Therefore $I=\int \limits_0^{\pi} \large\frac{(\pi-x) \sin x}{1+\cos ^2 x}$$dx$
$=\int \limits_0^{\pi} \large \frac{\pi \sin x }{1+\cos ^2 x }$$dx-\int \limits_0^{\pi} \large\frac{x \sin x}{1+\cos ^2 x}$$dx$
But $\large\frac{x \sin x}{1+\cos ^2 x}=$$I$
Therefore $I= \int \limits_0^\pi \large \frac{\pi \sin x }{1+\cos ^2 x} $$dx-I$
$2I=\pi \int \limits_0^\pi \large\frac{\sin x}{1+\cos ^2 x}$$dx$
Step 2:
Let $\cos x=t$ on differentiating w.r.t x,
$-\sin x dx= dt$
The limit also changes when we substitute t,
When $x=0,\; \cos 0=1=t$
When $ x=\pi, \;\cos \pi =-1 =t$
Therefore $ 2I=\pi \int \limits_{-1}^1 \large\frac{-dt}{1+t^2}$$=-\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} $$\tan ^{-1} (x)$
Therefore $2I= -\pi \int \limits_{-1}^1 \large\frac{dt}{1+t^2}$$=-\pi \bigg[\tan ^{-1}t\bigg]_{-1}^1$
Applying the limits we get,
$2I=-\pi \bigg[\tan ^{-1}(1)-\tan ^{-1} (-1)\bigg]$
$\tan ^{-1}=\pi/4 $ and $\tan ^{-1}(-1) =-\pi/4$
Therefore $ 2I= -\pi [\pi/4-(-\pi/4)]$
$=-\pi [\pi/4+\pi/4]$
$=-\pi \times (\pi/2)=-\large\frac{\pi^2}{2}$
answered May 6, 2013 by meena.p