Question

Evalaute: ${\int }_{-\pi /2}^{\pi /2}\sqrt{\cos x-{\cos }^{3}x}dx$

• A. $\frac{3}{4}$
• B. $-\frac{3}{4}$
• C. $\frac{4}{3}$
• D. $0$

Answer 1

The correct option is C $\frac{4}{3}$

${\int }_{-\pi /2}^{\pi /2}\sqrt{\cos x-{\cos }^{3}x}dx$

$={\int }_{-\pi /2}^{\pi /2}\sqrt{\cos x}|\sin x|dx$

Now, the given function is an even function:-

$=2{\int }_0^{\pi /2}\sqrt{\cos x}\sin xdx$ (since,$\sin x$ is positive in interval $(0,\frac{\pi }{2})$)

Let, $z=\cos x⟹dz=-\sin xdx$

And when$x=0,z=1$ and when $x=\frac{\pi }{2},z=0$

Hence, integration becomes:-

$2{\int }_1^0-\sqrt{z}dz={\int }_0^1\sqrt{z}dz$

$=2\times \frac{2}{3}{\left[z^{3/2}\right]}_0^1$

$=\frac{4}{3}$

Hence, answer is option-(C).