Mathematics - Definite Integration Question with Solution | TestHub

Given integral

$I={\int }_{-\pi /2}^{\pi /2}\sqrt{\left[\cos x\left(1-{\cos }^{2}x\right)\right]}dx={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\sqrt{\left(\cos x{\sin }^{2}x\right)}dx$.

$I={\int }_{-\pi /2}^{\pi /2}\sqrt{(\cos x)}|\sin x|dx$. Now,

$$|\sin x|=\{\begin{array}{c}-\sin x,\text{ if }-\pi /2\le x<0\\ \sin x,\text{ if }0<x\le \pi /2\end{array}$$

From (1) , we have

$I={\int }_{-\pi /2}^0\sqrt{(\cos x)}(-\sin x)dx+{\int }_0^{\pi /2}\sqrt{(\cos x)}\sin xdx$.

Putting $\cos x=t$, $-\sin xdx=dt$, we get

$I={\int }_0^1{t}^{1/2}dt-{\int }_1^0{t}^{1/2}dt=2{\int }_0^1{t}^{1/2}dt$.

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

$$=\frac{4}{3}$$