$\int\frac{\sqrt{\cot x}} {\sin x \cos x}dx$

Question

How can we evaluate the integral?: $$\int\frac{\sqrt{\cot x}} {\sin x \cos x}dx$$ To get an answer in terms of $\sqrt{\cot x}$? I tried using sin2x formula and then by-parts but didn't get the answer.

P.S- I am a beginner in integrals. Would love some tips on how to learn the same and what all to keep in mind while solving it's problems.

Answer 1

We have that $$\int\frac{\sqrt{\cot x}} {\sin x \cos x}dx=\int\frac{\sqrt{\cot x}} {\sin^2 x \cot x}dx=-\int\frac{d(\cot x)} {\sqrt{\cot x}}=-2\sqrt{\cot x}+C.$$

Answer 2

Hint...try substituting $u=\cot x$

Answer 3

$$ \begin{aligned} \int \frac{\sqrt{\cot x}}{\sin x \cos x} d x & =\int \frac{1}{\sin ^2 x \sqrt{\cot x}} d x \\ & =-\int \frac{d(\cot x)}{\sqrt{\cot x}} \\ & =-2 \sqrt{\cot x}+C \end{aligned} $$