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

Toolbox:
• $\int \limits _a^b f(x)dx=F(b)-F(a)$
• (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 $I=[0+1]+[-\frac{1}{3}]$

$=1-\frac{1}{3}=\frac{2}{3}$