The value of $I=\int_0^{\pi / 2} \frac{(\sin x+\cos x)^2}{\sqrt{1+\sin 2 x}} d x$ is

Question

Aicharya · Accepted Answer

$\int_0^{\frac{\pi}{2}} \frac{(\sin x+\cos x)^2}{\sqrt{(\sin x+\cos x)^2}} d x=\int_0^{\frac{\pi}{2}}(\sin x+\cos x) d x=|-\cos x+\sin x|_0^{\frac{\pi}{2}}=2$

The value of $I=\int_0^{\pi / 2} \frac{(\sin x+\cos x)^2}{\sqrt{1+\sin 2 x}} d x$ is

Solution