Question

${\int }_0^{\pi /2}\frac{\sin x+\cos x}{\sqrt{1+\sin 2x}}dx$ is equal to:

• A. $-\frac{\pi }{2}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{2}$
• D. $\pi$

Answer 1

The correct option is C $\frac{\pi }{2}$

Let $I={\int }_0^{\pi /2}\frac{\sin x+\cos x}{\sqrt{1+\sin 2x}}dx$
$={\int }_0^{\pi /2}\frac{\sin x+\cos x}{\sqrt{{\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x}}dx$
$={\int }_0^{\pi /2}\frac{(\sin x+\cos x)}{\sqrt{{(\sin x+\cos x)}^2}}dx$
$={\int }_0^{\pi /2}\frac{(\sin x+\cos x)}{(\sin x+\cos x)}dx={\int }_0^{\pi /2}1dx$
$={\left[x\right]}_0^{\pi /2}=\frac{\pi }{2}$