Question

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

• A. $2$
• B. $1$
• C. $3$
• D. $0$

Answer 1

The correct option is A $2$

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

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

$={\int }_0^{\frac{\pi }{2}}\sin xdx+{\int }_0^{\frac{\pi }{2}}\cos xdx$

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

$=-\cos \frac{\pi }{2}+\cos 0+\sin \frac{\pi }{2}-\sin 0$

$=1+1=2$