Question

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

Answer 1

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

We are given that $I=\int \frac{sinx-cosx}{\sqrt{sin2x}}dx$

now, divided $cosx$ in Numerator &

Denominator

$I=\int \frac{tanx-1}{\frac{2sinx.cosx}{co{s}^{2}x}}dx$

$=\int \frac{tanx-1}{\sqrt{2tan}}dx$

$=\frac{1}{\sqrt{2}}\int \left(\sqrt{tanx-\sqrt{cotx}}\right)dx$

$=\frac{1}{\sqrt{2}}\int \sqrt{tanx}dx+\frac{1}{\sqrt{2}}\int -\sqrt{cotx}dx$

Let $\sqrt{tanx}=t\Rightarrow tanx=t^2$

take differentiation wrt x. we get,

$\therefore se{c}^{2}xdx=2tdt$

$dx=\frac{2tdt}{se{c}^{2}x}=\frac{2tdt}{1+ta{n}^{2}x}$

$\therefore dx=\frac{2tdt}{1+t^2}$

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

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

$=\frac{2}{\sqrt{2}}[\int \frac{1+t^2-1}{1+t^2}dt]-\frac{\left[2ta{n}^{-1}t\right]}{\sqrt{2}}$

$=\sqrt{2}[t-ta{n}^{-1}t]-\sqrt{2}ta{n}^{-1}t+c$

$=\sqrt{2}t-2\sqrt{2}ta{n}^{-1}t+c$

$I=\sqrt{2tanx}-2\sqrt{2}ta{n}^{-1}\left(\sqrt{tanx}\right)+c$