Question

${\int }_0^{100\pi /2}\sqrt{1+\cos 2x}dx$ is equal to

• A. $0$
• B. $100\sqrt{2}$
• C. $200\sqrt{2}$
• D. $100$

Answer 1

The correct option is B $100\sqrt{2}$

Let $I=\int \sqrt{\cos 2x+1}dx$
$u=2x\Rightarrow du=2dx$

$\therefore I=\frac{1}{2}\int \sqrt{\cos u+1}dx=\frac{1}{2}\int \sqrt{2}\sqrt{\frac{{\cos }^{2}u}{2}}du$

$=\frac{1}{2}\int \sqrt{\frac{{\cos }^{2}u}{2}}du$
Substitute $t=\frac{u}{2}\Rightarrow dt=\frac{du}{2}$

$I=\sqrt{2}\int \sqrt{{\cos }^{2}t}dt=\sqrt{2}\int \cos tdt$

$=\sqrt{2}\sin t=\sqrt{2}\sin \left(\frac{u}{2}\right)=\sqrt{2}\sin x$

$\therefore {\int }_0^{100\pi }\sqrt{1+\cos 2x}dx={\left[\sqrt{2}\sin x\right]}_0^{100\frac{\pi }{2}}=100\sqrt{2}$