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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.

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

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  • $\begingroup$ Thanks a lot..... Really appreciate it👍 $\endgroup$ – Hani Mohammed Sep 4 '16 at 12:57
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Hint...try substituting $u=\cot x$

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