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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Evaluate the following: $\int\limits_0^\pi x\sin x\cos^2xdx$

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Toolbox:
  • Method of integration by parts
  • $\int udv=uv-\int vdu$
  • $\sin ^3x=3\sin x-4 \sin ^3x$
Step 1:
Let $I=\int\limits_0^\pi x\sin x\;\cos^2xdx$
$\cos ^2 x=1-\sin ^2x$
Hence $\int\limits_0^\pi x\sin x(1-\sin ^2x)dx$
$=\int\limits_0^\pi (x\sin x-x \sin ^3 x) dx$
$=\int\limits_0^\pi x\sin x dx-\int\limits_0^\pi x\sin ^3x dx$
$\sin 3x=3 \sin x-4 \sin ^3 x$
Therefore $sin ^3 x=\large\frac{3 \sin x-\sin 3x}{4}$
$I=\int\limits_0^\pi x\sin x dx- \int\limits_0^\pi x \bigg[\large\frac{3 \sin x-\sin 3x}{4}\bigg]$$dx$
On simplifying we get,
$=\int\limits_0^\pi x\sin x dx-\frac{3}{4}\int\limits_0^\pi x\sin x +\frac{1}{4} \int\limits_0^\pi x\sin 3x dx$
$I=\large\frac{1}{4}$$\int\limits_0^\pi x\sin x dx+\frac{1}{4}\int\limits_0^\pi x\sin 3x dx$
Step 2:
Let us take as $I=\large\frac{1}{4}$$ [I_1+I_2]$
$I_1=\int\limits_0^\pi x\sin x dx$
This can be solved by the method of integration by parts
$\int udv=uv-\int vdu$
Here let $u=x,$ on differentiating with respect to x we get, $du=dx$
Let $dv=\sin x dx$ , on integrating we get $v=-\cos x$
Now substituting for u,v,du and dv we get,
$\int\limits_0^\pi x\sin x dx=x(-\cos x)-\int\limits_0^\pi -\cos x dx$
$=-(x \cos x)_0^{\pi}+\int\limits_0^\pi \cos x dx$
On integrating we get,
$I_1=-[x \cos x]_0^{\pi}+[\sin x]_0^{\pi}$
On applying limits we get
$I_1=-[\pi. \cos \pi -0]+[\sin \pi -\sin 0]$
$\sin \pi=0\;and\;\cos \pi=-1\;and\; \sin 0=0$
Therefore $I_1=-[\pi(-1)]+0$
$I_1=\pi$
Step 3:
Consider $I_2 =\int\limits_0^\pi x \sin 3x dx$
Similarly, let us solve this by the method of integration.
Let $u=x,$ on differentiating we get $du=dx$
Let $dv=\sin 3xdx$, on integrating we get $v=\frac{-1}{3} \cos 3x$
Now substituting for u,v,du and dv we get,
$\int\limits_0^\pi x\sin 3x dx=x(-\frac{1}{3}\cos x)_0^{\pi}+\frac{1}{3}\int\limits_0^\pi \cos 3x dx$
On integrating we get,
$=\large-\frac{1}{3}$$[x \cos 3x]_0^{\pi}+$$\large\frac{1}{3}[\frac{1}{3}$$\sin 3x]_0^{\pi}$
$=\large-\frac{1}{3}$$[x \cos 3x]_0^{\pi}+$$\large\frac{1}{9}$$[\sin 3x]_0^{\pi}$
On applying limits we get,
$I_2=-\frac{1}{3} [\pi \cos 3 \pi -0]+\frac{1}{9}[\sin 3 \pi-\sin 0]$
$\sin 3\pi=\sin 0=0$ and $\cos 3 \pi =-1$
Therefore $I_2 =\large-\frac{1}{3}$$[\pi(-1)]+0$
$=\large\frac{1}{3} \pi$
Step 4:
Therefore $ I=\large\frac{1}{4}$$[I_1+I_2]$
$=\large\frac{1}{4}[\pi+\frac{1}{3}\pi]=\frac{1}{4}.\frac{4\pi}{3}$
$=\large\frac{\pi}{3}$
answered Apr 19, 2013 by meena.p
 
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