Question

Solve:

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

Answer 1

$I=\int \sqrt{\frac{\cos x}{\sin x}}dx=\int \sqrt{\cot x}dxLet,\cot x=u\Rightarrow x=arccotu\therefore I=\int -\frac{\sqrt{u}}{1+u^2}du\Rightarrow I=-\int \frac{\sqrt{u}}{1+u^2}duagain,applyingu=\frac{v}{2}wegetI=-\int \frac{\sqrt{2}\sqrt{v}}{4+v^2}dv\Rightarrow I=-\sqrt{2}\int \frac{\sqrt{v}}{4+v^2}again,let,w=\sqrt{v}\Rightarrow I=-\sqrt{2}\int \frac{2w^2}{4+w^4}dw\Rightarrow I=-\sqrt{2}.2\int \frac{w}{4(w^2-2w+2)}-\frac{w}{4(w^2+2w+2)}dw\left[bypartialfraction\right]\Rightarrow I=-\sqrt{2}.2[\int \frac{w}{4(w^2-2w+2)}dw-\int \frac{w}{4(w^2+2w+2)}dw]=-\sqrt{2}.2[[\frac{1}{8}log(\left[{\left(\sqrt{2\cot \left(x\right)}\right)}^2\right]-2\sqrt{cotx}+2)+2\arctan (\sqrt{2cotx}-1)]-[\frac{1}{8}log(\left[{\left(\sqrt{2\cot \left(x\right)}\right)}^2\right]+2\sqrt{cotx}+2)-2\arctan (\sqrt{2cotx}+1)]]+C$