Prove the following Exercise: ^ 2 _ 0 ^ 3 x dx=2 3 - Vidyadip

Question

Prove the following Exercise:
$\int^{\frac{\pi}{2}}_{0}\sin^{3}\text{x dx}=\frac{2}{3}$

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Answer

$\text{Let I}=\int^{\frac{\pi}{2}}_{0}\sin^{3}\text{x dx}$
$\text{I}=\int^{\frac{\pi}{2}}_{0}\sin^{2}\text{x.}\sin\text{x dx}$
$=\int^{\frac{\pi}{2}}_{0}\Big(1-\cos^{2}\text{x}\Big)\sin\text{x dx}$
$=\int^{\frac{\pi}{2}}_{0}\sin\text{x dx}-\int^{\frac{\pi}{2}}_{0}\cos^{2}\text{x}.\sin\text{x dx}$
$=\Big[-\cos\text{x}\Big]^{\frac{\pi}{0}}_{0}-\Big[\frac{\cos^{3}\text{x}}{3}\Big]^{\frac{\pi}{2}}_{0}$
$=1+\frac{1}{3}[-1]=1-\frac{1}{3}=\frac{2}{3}$
Hence, the given result is Proved.

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