Question

The value of ∫{(sinx + cosx)^2/√(1 + sin2x)}dx where,x→(0, π/2) is.

Answer 1

We have to evaluate the definite integral

$\int\limits_0^{\pi/2} {\frac{(\sin x+\cos x)^2}{\sqrt{1+\sin 2x}}} \, dx$.

Now note that

$1+\sin 2x=\sin^2 x+\cos ^2 x+2\sin x \cos x\\ 1+\sin 2x=(\sin x + \cos x)^2\\ \sqrt{1+\sin 2x}=\sin x + \cos x$

The integrand is

$\frac{(\sin x+\cos x)^2}{\sqrt{1+\sin 2x}}=\frac{(\sin x+\cos x)^2}{\sin x+\cos x}=\sin x+\cos x$

The integral becomes,

$\int\limits_0^{\pi/2} {\frac{(\sin x+\cos x)^2}{\sqrt{1+\sin 2x}}} \, dx = \int\limits_0^{\pi/2} {(\sin x+\cos x)} \, dx \\ \int\limits_0^{\pi/2} {\frac{(\sin x+\cos x)^2}{\sqrt{1+\sin 2x}}} \, dx = [\sin x-\cos x]_0^{\pi/2}\\</p><p> \int\limits_0^{\pi/2} {\frac{(\sin x+\cos x)^2}{\sqrt{1+\sin 2x}}} \, dx = 1-0-(0-1) \int\limits_0^{\pi/2} {\frac{(\sin x+\cos x)^2}{\sqrt{1+\sin 2x}}} \, dx =2$