Question

Solve:

${\int }_0^{\pi /2}\sqrt{1+sin2x}dx=$

• A. $1/2$
• B. $1$
• C. $2$
• D. $\frac{3}{2}$

Answer 1

Answer: (C)

$2$

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

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

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

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

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

$=[-(\cos \frac{\pi }{2}-\cos 0)+(\sin \frac{\pi }{2}-\sin 0)]$

$=[-(0-1)+(1-0)]$

$=1+1=2$