# Evaluate the following integrals:

Question:

$\int \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x$

Solution:

$\int \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x$

We can write above integral as

$=\int \frac{\sin x+\cos x}{\sqrt{1-1+\sin 2 x}} d x$ [Adding and subtracting 1 in denominator]

$=\int \frac{\sin x+\cos x}{\sqrt{1-(1-\sin 2 x)}} d x$

$=\int \frac{\sin x+\cos x}{\sqrt{1-\left(\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x\right)}} d x \because \sin ^{2} x+\cos ^{2} x=1$ and

$\sin 2 x=2 \sin x \cos x$

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

Put $\sin x-\cos x=t$

Differentiating w.r.t $x$ we get,

$(\cos x+\sin x) d x=d t$

Putting values we get,

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

$=\int \frac{d t}{\sqrt{1-t^{2}}}=\sin ^{-1} t+C$

Putting value of $t$ we get,

$\therefore \int \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x=\sin ^{-1}(\sin x-\cos x)+C$