Question

The value of integral ${\int }_0^{500}\sqrt{1+\cos 2x}dx$ is

• A. $\sqrt{2}sin500$
• B. $100\sqrt{2}$
• C. $200\sqrt{2}$
• D. $200$

Answer 1

Answer: (A)

$\sqrt{2}sin500$

$I={\int }_0^{500}\sqrt{1+\cos 2x}dx$

$={\int }_0^{500}\sqrt{1+2{\cos }^{2}x-1}dx$

$=\sqrt{2}{\left(sinx\right)}_0^{500}$

$=\sqrt{2}\sin 500$