Question

Integrate dx/(1-sin2x)^1/2

Answer 1

Solution :

$\therefore\: \mathrm{\int \frac{dx}{\sqrt{1-sin2x}}}$

$\mathrm{=\int \frac{dx}{\sqrt{sin^{2}x+cos^{2}x-2\:sinx\:cosx}}}$

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

$\mathrm{=\int \frac{dx}{sinx-cosx}}$

$\mathrm{=\sqrt{2}\int \frac{dx}{\frac{1}{\sqrt{2}}sinx-\frac{1}{\sqrt{2}}cosx}dx}$

$\mathrm{=\sqrt{2}\int \frac{dx}{sinx\:cos\frac{\pi}{4}-cosx\:sin\frac{\pi}{4}}}$

$\mathrm{=\sqrt{2}\int \frac{dx}{sin(x-\frac{\pi}{4})}}$

$\mathrm{=\sqrt{2}\int cosec(x-\frac{\pi}{4})\:dx}$

$\mathrm{=\sqrt{2}\:log|tan\frac{(x-\frac{\pi}{4})}{2}|+C}$ ,

where C is constant of integration

$\mathrm{=\sqrt{2}\:log|tan(\frac{x}{2}-\frac{\pi}{8})|+C}$ ,

which is the required integral. (Ans.)