Mathematics

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


ANSWER

$$\dfrac{2}{3}$$


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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