Mathematics

# Solve $\int_{0}^{\frac{\pi }{2}}{{{\sin }^{3}}xdx}$

$\dfrac{2}{3}$

##### SOLUTION

Consider the given integral.

$I=\int_{0}^{\frac{\pi }{2}}{{{\sin }^{3}}xdx}$

We know that

$\sin 3\theta =3\sin \theta -4si{{n}^{3}}\theta$

${{\sin }^{3}}\theta =\dfrac{3\sin \theta -sin3\theta }{4}$

Therefore,

$I=\int_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{3\sin x-sin3x}{4} \right)dx}$

$I=\dfrac{1}{4}\left( -3\cos x-\left( \dfrac{-\cos 3x}{3} \right) \right)_{0}^{\dfrac{\pi }{2}}$

$I=\dfrac{1}{4}\left( -3\cos x+\dfrac{\cos 3x}{3} \right)_{0}^{\dfrac{\pi }{2}}$

$I=\dfrac{1}{4}\left( \left( -3\cos \dfrac{\pi }{2}+\dfrac{\cos \dfrac{3\pi }{2}}{3} \right)-\left( -3\cos 0+\dfrac{\cos 0}{3} \right) \right)$

$I=\dfrac{1}{4}\left( \left( 0+\dfrac{0}{3} \right)-\left( -3+\dfrac{1}{3} \right) \right)$

$I=\dfrac{1}{4}\left( \dfrac{8}{3} \right)=\dfrac{2}{3}$

Hence, this is the answer.

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
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Chapters 126
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