Question

The value of the definite integral $I=\underset{0}{\overset{\pi }{\int }}x\sqrt{1+\left|\cos x\right|}dx$ is equal to

• A. $2\sqrt{2}\pi$
• B. $\sqrt{2}\pi$
• C. $2\pi$
• D. $4\pi$

Answer 1

Answer: (C)

$2\pi$

$I={\int }_0^{\pi }x\sqrt{1+|\cos x|}dx$

Since, $\cos x$ is negative in $\frac{\pi }{2}\le x\le \pi$
${\int }_0^{\pi /2}x\sqrt{1+\cos x}dx+{\int }_{\pi /2}^{\pi }x\sqrt{1-\cos x}dx$
we know that
$1+\cos \frac{2x}{2}=2{\cos }^2\frac{x}{2}and1-\cos \frac{2x}{2}=2{\sin }^2\frac{x}{2}$

Hence,

${\int }_0^{\pi /2}x\sqrt{2{\cos }^2\frac{x}{2}}dx+{\int }_{\pi /2}^{\pi }x\sqrt{2{\sin }^2\frac{x}{2}}dx$

${\int }_0^{\pi /2}\sqrt{2}x\cos \frac{x}{2}dx+{\int }_{\pi /2}^{\pi }\sqrt{2}x\sin \frac{x}{2}dx$

integrating by parts

$\sqrt{2}[x\int \cos \frac{x}{2}dx-\int \frac{dx}{dx}\int \cos \frac{x}{2}dxdx{]}_0^{\pi /2}$$+\sqrt{2}[x\int \sin \frac{x}{2}dx-\int \frac{dx}{dx}\int \sin \frac{x}{2}dxdx{]}_{\pi /2}^{\pi }$

$\sqrt{2}[2x\sin \frac{x}{2}-2\int \sin \frac{x}{2}dx{]}_0^{\pi /2}+$$\sqrt{2}[-2x\cos \frac{x}{2}+2\int \cos \frac{x}{2}dx{]}_{\pi /2}^{\pi }$

$2\sqrt{2}[x\sin \frac{x}{2}+2\cos \frac{x}{2}{]}_0^{\pi /2}+$$2\sqrt{2}[-x\cos \frac{x}{2}+2\sin \frac{x}{2}{]}_{\pi /2}^{\pi }$

$2\sqrt{2}[\frac{\pi }{2}\sin \frac{\pi }{4}+2\cos \frac{\pi }{4}-0\sin \left(0\right)-2\cos \left(0\right)]$$+2\sqrt{2}[-\pi \cos \frac{\pi }{2}+2\sin \frac{\pi }{2}+\frac{\pi }{2}\cos \frac{\pi }{4}-2\sin \frac{\pi }{4}]$

$2\sqrt{2}[\frac{\pi }{2\sqrt{2}}+\frac{2}{\sqrt{2}}+0-2+0+2+\frac{\pi }{2\sqrt{2}}-\frac{2}{\sqrt{2}}]$

$2\sqrt{2}[\frac{\pi }{2\sqrt{2}}+\frac{\pi }{2\sqrt{2}}]$

$2\sqrt{2}[\frac{2\pi }{2\sqrt{2}}]$

$2\pi$