Question

${\int }_0^{\pi }\sqrt{|\cos x|-|{\cos }^{3}x|}dx$ is?

• A. $\frac{4}{3}$
• B. $1$
• C. $3$
• D. $4$

Answer 1

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

Given : ${\int }_0^{\pi }\sqrt{\left|\cos x\right|-\left|{\cos }^{3}x\right|dx}$

$I={\int }_0^{\pi }\sqrt{\left|\cos x\right|-\left|{\cos }^{3}x\right|dx}={\int }_0^{\pi }\sqrt{\left|\cos x\right||1-{\cos }^{2}x|dx}I={\int }_0^{\pi }\sqrt{\left|\cos x\right|}\left|\sin x\right|dxI={\int }_0^{\pi }\sqrt{\left|\cos x\right|}\left(\sin x\right)dx\sin x>0(0<x<\pi )\cos x=tx\to 0t\to 1dt=-\sin xdxx\to \pi t\to -1I=-{\int }_1^{-1}\sqrt{\left|t\right|}dt={\int }_{-1}^1\sqrt{\left|t\right|}dt={\int }_{-1}^0\sqrt{-t}dt+{\int }_0^1\sqrt{t}dtI=-\frac{2}{3}{\left|{(-t)}^{\frac{3}{2}}\right|}_{-1}^0+\frac{2}{3}{\left|t^{\frac{3}{2}}\right|}_0^{1}I=-\frac{2}{3}[0-1]+\frac{2}{3}(1-0)=\frac{4}{3}$

Hence the correct answer is $\frac{4}{3}$