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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the definite integrals\[\int\limits_0^\frac{\pi}{2}\sin 2x\tan^{-1}(sin x)\;dx\]

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  • (i)$\int \limits_a^b f(x)dx=F(b)-F(a)$
  • (ii) In a integral function, $\int f(x)dx, let f(x)=t,\; then\; f'(x)dx=dt,$ hence $ f(x)dx=\int t dt$
  • (iii)$\int \frac{dx}{a^2+x^2}=\frac{1}{a}\tan ^{-1}(x/a)+c$
  • (iv)$\sin 2x=2\sin x\cos x$
  • (v)$\int udv=uv-\int vdu$
Given $I=\int\limits_0^\frac{\pi}{2}\sin 2x\tan^{-1}(sin x)\;dx$
we know $\sin 2x=2\sin x\cos x$
Hence $I=\int \limits_0^{\pi/2} 2\sin x \cos x \tan^{-1}(\sin x)dx$
Now, let $\sin x=t,$ on differentiating w.r.t x we get
$\cos x dx=dt$
The limit also changes when su bstitute for t,
When $x=0,\sin x=0 \qquad t=0$
When $ x=\pi/2,\sin \pi/2=1 \qquad t=1$
on substituting t and dt
Therefore $I=2\int \limits_0^1 t.\tan^{-1}(t)dt$
This is the form $\int udv=uv-\int vdu$
Let $u=\tan^{-1}t$ on differentiating w.r.t to t
Let $dv=tdt$ on integrating we get,
Now substituting for u,v,du and dt,
$I=2 \bigg\{[\tan ^{-1}t.\frac{t^2}{2}]_0^1=\int \limits_0^1 \frac{t^2}{2}.\frac{1}{1+t^2}.dt\bigg\}$
$=2 \bigg\{[\frac{t^2}{2}\tan ^{-1}t]_0^1-\frac{1}{2}\int \limits_0^1 \frac{t^2}{1+t^2}.dt\bigg\}$
Consider $I_1=\int \frac {t^2}{1+t^2} dt$
Add and subtract +1 and -1 to the numerator,
$I-1=\int \limits_0^1 \frac{t^2+t-1}{1+t^2}dt$
Seperating the terms we get,
$\int \limits_0^1 \frac{t^2+1}{1+t^2}dt-\int \limits_0^1\frac{1}{1+t^2}dt$
$=\int \limits_0^1dt-\int \limits_0^1 \frac{dt}{1+t^2}$
$\int \frac{dt}{1+t^2}$ is of the form $\int \frac{dx}{a^2+x^2}=\frac{1}{a}\tan ^{-1}(x/a)+c$
Therefore on integrating we get,
$[t]_0^1-[\tan -1 (t)]_0^1$
Hence $I=2\bigg\{[\frac{t^2}{2} \tan^{-1} t]_0^1-\frac{1}{2}\bigg\{[t]_0^1+[\tan ^{-1}(t)]_0^1 \bigg\} \bigg\}$
On applying limits we get ,
$2\bigg\{\frac{1}{2} \tan^{-1} (1)-0]-\frac{1}{2}[1-0]+\frac{1}{2}[\tan ^{-1}(1)-0]\bigg\} $
$I=2\bigg\{\frac{1}{2} (\tan^{-1} (1)-\frac{1}{2}+\frac{1}{2}\tan ^{-1}(1)\bigg\} $
But $\tan^{-1}(1)=\pi/4$



answered Feb 19, 2013 by meena.p

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