Question

$\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{\sin x+\cos x}{\sqrt{\sin 2x}}dx$ is equal to

• A. $0$
• B. $1$
• C. $\frac{\pi }{2}$
• D. $\pi$

Answer 1

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

$\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{\sin x+\cos x}{\sqrt{1-(1-\sin 2x)}}dx=\underset{0}{\overset{\frac{\pi }{4}}{\int }}\frac{\sin x+\cos x}{\sqrt{1-(\sin x-\cos x{)}^2}}dx$
Let $(\sin x-\cos x)=t\Rightarrow dt=\cos x+\sin x)dx$$=\underset{-1}{\overset{0}{\int }}\frac{dt}{\sqrt{1-t^2}}=[{\sin }^{-1}t{]}_{-1}^0$
$=[{\sin }^{-1}\left(0\right)-{\sin }^{-1}(-1)]=\frac{\pi }{2}$