Question

Evaluate:${\int }_0^{\frac{\pi }{2}}{\cos }^{3}xdx$

Answer 1

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

$=\frac{1}{4}{\int }_0^{\frac{\pi }{2}}4{\cos }^{3}xdx$

We know that $\cos 3x=4{\cos }^{3}x-3\cos x\Rightarrow 4{\cos }^{3}x=3\cos x+\cos 3x$

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

$=\frac{1}{4}{\int }_0^{\frac{\pi }{2}}3\cos xdx+\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\cos 3xdx$

$=\frac{3}{4}{\int }_0^{\frac{\pi }{2}}\cos xdx+\frac{1}{4}{\int }_0^{\frac{\pi }{2}}\cos 3xdx$

$=\frac{3}{4}{\left[\sin x\right]}_0^{\frac{\pi }{2}}+\frac{1}{4}{\left[\frac{\sin 3x}{3}\right]}_0^{\frac{\pi }{2}}$

$=\frac{3}{4}[\sin \frac{\pi }{2}-\sin 0]+\frac{1}{12}[\sin \frac{3\pi }{2}-\sin 0]$

$=\frac{3}{4}[1-0]+\frac{1}{12}[\sin (\pi +\frac{\pi }{2})-0]$

$=\frac{3}{4}+\frac{1}{12}[-\sin \frac{\pi }{2}-0]$

$=\frac{3}{4}-\frac{1}{12}[1-0]$

$=\frac{3}{4}-\frac{1}{12}$

$=\frac{3\times 3}{4\times 3}-\frac{1}{12}$

$=\frac{9}{12}-\frac{1}{12}$

$=\frac{8}{12}=\frac{2}{3}$