Question

$\underset{0}{\overset{\pi }{\int }}x\sqrt{1+\left|\cos x\right|}dx$ is equal to

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

Answer 1

The correct option is C $2\pi$

Let,
$I=\underset{0}{\overset{\pi }{\int }}x\sqrt{1+\left|\cos x\right|}dx\Rightarrow I=\underset{0}{\overset{\pi }{\int }}(\pi -x)\sqrt{1+\left|\cos x\right|}dx\left[\text{using}\underset{a}{\overset{b}{\int }}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx\right]$
Adding both, we get
$\Rightarrow 2I=\pi \underset{0}{\overset{\pi }{\int }}\sqrt{1+\left|\cos x\right|}dx\Rightarrow 2I=2\pi \underset{0}{\overset{\frac{\pi }{2}}{\int }}\sqrt{1+\cos x}dx[\because \underset{0}{\overset{2a}{\int }}f\left(x\right)dx=2\underset{0}{\overset{a}{\int }}f\left(x\right)dx,\text{when}f\left(x\right)=f(2a-x)]\Rightarrow I=\pi \sqrt{2}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos \frac{x}{2}dx\Rightarrow I=2\sqrt{2}\pi \cdot {\left[\sin \frac{x}{2}\right]}_0^{\frac{\pi }{2}}=2\pi$