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Prove the following\[\int\limits_0^\frac{\pi}{2}\sin^3x\;dx=\frac{2}{3}\]

1 Answer

  • $\int \limits _a^b f(x)dx=F(b)-F(a)$
  • (ii) $ \sin ^2 x=1-\cos ^2 x$
  • (iii)$ \int \sin x dx=-\cos x +c$
Given $I=\int \limits_0^{\pi/2} \sin ^x dx=\frac{2}{3}$
$\sin ^3 x$ can be written as $\sin ^2x.\sin x$
Hence $I=\int \limits_0 ^ {\pi/2}sin ^2x.\sin x dx$
But $ \sin ^2x=1-\cos ^2 x$
Therefore $I=\int \limits_0^{\pi/2} (1-\cos ^2 x).\sin x dx$
on expanding we get
$I=\int \limits_0^{\pi/2} (\sin x -\cos ^2x.\sin x) dx$
on seperating the terms,
$I=\int \limits_0^{\pi/2} \sin x dx -\int \limits _0^ {\pi/2} \cos ^2 x.\sin x dx$
Let $I_1=\int \limits_0^{\pi/2} \sin x dx$
Consider $I_2=\int \limits _0^ {\pi/2} \cos ^2 x.\sin x dx$
Let $ \cos x =t,$ on differentiating w.r.t x we get
$-\sin x dx =dt \qquad => \sin x dx =-dt$
The value of the limit changes when we substitute t
When x=0 $\cos 0=1$ hence t=1 and When $x=\frac{\pi}{2}, \cos \pi/2=0,hence\; t=0$
Therefore $I_2=-\int\limits_1^0 t^2 dt$
Therefore $I=I_1+I_2$
$=\int \limits_0^{\pi/2} \sin x dx-\int\limits_0^1 -t^2 dt$
On integrating we get,
$I=[-\cos x ]_0^{\pi/2}+[\frac{t^3}{3}]_1^0$
On applying the limits we get ,
$I=[-\cos \pi/2 -(-\cos 0)]+[0-\frac{1^3}{3}]$
But $\cos \pi/2=0\; and\; \cos 0=1$
Hence $I=[0+1]+[-\frac{1}{3}]$



answered Mar 13, 2013 by meena.p
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