Question

The value of $I={\int }_0^{\pi /2}\frac{(sinx+cosx{)}^2}{\sqrt{1+\sin 2x}}$ dx is:

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is C $2$

${\int }_0^{\frac{\pi }{2}}\frac{{(\sin x+\cos x)}^2}{\sqrt{1+\sin 2x}}dx={\int }_0^{\frac{\pi }{2}}\frac{{(\sin x+\cos x)}^2}{\sqrt{{\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x}}dx$
$={\int }_0^{\frac{\pi }{2}}\frac{{(\sin x+\cos x)}^2}{\sqrt{{(\sin x+\cos x)}^2}}dx={\int }_0^{\frac{\pi }{2}}(\sin x+\cos x)dx={[-\cos x+\sin x]}_0^{\frac{\pi }{2}}=-0+1+1-0=2$