Question

Solve ${\int }_0^{\frac{\pi }{2}}{\sin }^{3}xdx$

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{5}{3}$
• D. $-\frac{2}{3}$

Answer 1

The correct option is B $\frac{2}{3}$

Consider the given integral.

$I={\int }_0^{\frac{\pi }{2}}{\sin }^{3}xdx$

We know that

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

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

Therefore,

$I={\int }_0^{\frac{\pi }{2}}\left(\frac{3\sin x-sin3x}{4}\right)dx$

$I=\frac{1}{4}{(-3\cos x-\left(\frac{-\cos 3x}{3}\right))}_0^{\frac{\pi }{2}}$

$I=\frac{1}{4}{(-3\cos x+\frac{\cos 3x}{3})}_0^{\frac{\pi }{2}}$

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

$I=\frac{1}{4}((0+\frac{0}{3})-(-3+\frac{1}{3}))$

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

Hence, this is the answer.