Question

The value of ${\int }_0^{\frac{\pi }{2}}\sqrt{1+\sin 2x}dx$ is:

• A. $1$
• B. $-2$
• C. $\sqrt{2}$
• D. $2$

Answer 1

Answer: (D)

$2$

${\int }_0^{\frac{\pi }{2}}\sqrt{1+\sin 2x}dx={\int }_0^{\frac{\pi }{2}}\sqrt{{\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x}dx={\int }_0^{\frac{\pi }{2}}\sqrt{{(\sin x+\cos x)}^2}dx={\int }_0^{\frac{\pi }{2}}(\sin x+\cos x)dx$

$=[-\cos x+\sin x{]}_0^{\frac{\pi }{2}}=1+1=2$