Question

Evaluate: $\int \frac{cosx+sinx}{\sqrt{sin2x}}dx$

Answer 1

$I=\int \frac{cosx+sinx}{\sqrt{sin2x}}dx$

$=\int \frac{cosx+sinx}{\sqrt{1+sin2x-1}}dx$

$=\int \frac{cosx+sinx}{\sqrt{1-(1-sin2x)}}dx$

$=\int \frac{cosx+sinx}{\sqrt{1-(si{n}^{2}x+co{s}^{2}x-2sinxcosx)}}dx$

$=\int \frac{(cosx+sinx)}{\sqrt{1-(sinx-cosx{)}^2}}dx$

put $sinx-cosx=t$

$\Rightarrow (cosx+sinx)dx=dt$

$=\int \frac{dt}{\sqrt{1-t^2}}$

$=si{n}^{-1}t+c$

$=si{n}^{-1}(sinx-cosx)+c$