Question

The value of ${\int }_0^{\frac{\pi }{4}}{\cos }^{3}2xdx$ is:

• A. $\frac{2}{3}$
• B. $\frac{1}{3}$
• C. $0$
• D. $\frac{2\pi }{3}$

Answer 1

The correct option is B $\frac{1}{3}$

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

Using $\cos 3x=4{\cos }^{3}x-3\cos x$

$\Rightarrow {\cos }^{3}x=\frac{1}{4}(\cos 3x+3\cos x)$

So ${\cos }^{3}2x=\frac{1}{4}(\cos 6x+3\cos 2x)$

Therefore,

$I=\frac{1}{4}{\int }_0^{\frac{\pi }{4}}(\cos 6x+3\cos 2x)dx$

$=\frac{1}{4}{[\frac{1}{6}\sin 6x+\frac{3}{2}\sin 2x]}_0^{\frac{\pi }{4}}$

$=\frac{1}{4}[\frac{1}{6}\sin \frac{3\pi }{2}+\frac{3}{2}\sin \frac{\pi }{2}]$

$=\frac{1}{4}(-\frac{1}{6}+\frac{3}{2})=\frac{1}{4}\cdot \frac{8}{6}=\frac{1}{3}$