Question

Evaluate:

$\int \frac{\cos x-\sin x}{\sqrt{\sin 2x}}dx$.

Answer 1

Now,

$\int \frac{\cos x-\sin x}{\sqrt{\sin 2x}}dx$

$=\int \frac{\cos x-\sin x}{\sqrt{1+\sin 2x-1}}dx$

$=\int \frac{\cos x-\sin x}{\sqrt{{\sin }^{2}x+{\cos }^{2}x+2\sin x.\cos x-1}}dx$ [ Since ${\sin }^{2}x+{\cos }^{2}x=1$]

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

$=\log |(\sin x+\cos x)+\sqrt{(\sin x+\cos x{)}^2-1}|+c$ [ Where $c$ is integrating constant]

$=\log |(\sin x+\cos x)+\sqrt{\sin 2x}|+c$