Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is C $2$

${\int }_0^{\frac{\pi }{2}}\frac{(sinx+cosx{)}^2}{\sqrt{1+sin2x}}dx={\int }_0^{\frac{\pi }{2}}(sinx+cosx)dx$
$={\int }_0^{\frac{\pi }{2}}sinxdx+{\int }_0^{\frac{\pi }{2}}cosxdx$
$1+1=2$