Mathematics

# Evaluate: $\displaystyle \int_{0}^{\pi /2}\sin^3 x\ dx$

##### SOLUTION

$I=\displaystyle \int_{0}^{\pi /2} \sin^3 x\ dx$

$I=\dfrac {1}{4} \displaystyle \int_{0}^{\pi /2} (3\sin x-\sin 3x)dx$

$I=\dfrac {1}{4} \left [-3\cos x+\dfrac {1}{3} \cos 3x \right]_0^{\pi /2}$

$\ I=\dfrac {1}{4} \left [\left (-3\cos \dfrac {\pi}{2} +\dfrac {1}{3} \cos \dfrac {3\pi}{2}\right)-\left (-3+\dfrac {1}{3}\right) \right]$

$I=\dfrac {2}{3}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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