(a) Find: ∫ (√(cot x))/(sin x cos x) d x

(a)Let $\displaystyle \cot x = t$, then $\displaystyle -\operatorname{cosec}^2 x\,dx = dt$$$\therefore \int \frac{\sqrt{\cot x}}{\sin x \cos x}\,dx = \int \frac{\sqrt{\cot x}}{\cot x}\operatorname{cosec}^2 x\,dx = -\int \frac{\sqrt{t}}{t}\,dt = -\int \frac{1}{\sqrt{t}}\,dt$$$$= -2\sqrt{t} + C$$$$= -2\sqrt{\cot x} + C$$Or(b)Let $\displaystyle I = \displaystyle\int \frac{1}{x(x^2+4)}\,dx = \int \frac{1}{x^3\left(1+\dfrac{4}{x^2}\right)}\,dx$Let $\displaystyle 1 + \dfrac{4}{x^2} = t \Rightarrow \dfrac{-8}{x^3}\,dx = dt$$$\therefore\ I = \frac{-1}{8}\int \frac{dt}{t}$$$$= \frac{-1}{8}\log|t| + C$$$$= \frac{-1}{8}\log\left[1 + \frac{4}{x^2}\right] + C$$