Question

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

• A. $2$
• B. $1$
• C. $0$
• D. $3$

Answer 1

The correct option is A $2$

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

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

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

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

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

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

$=(0-(-1\left)\right)+(1-0)$

$=2$

Hence, answer is option-(A).