(a)OR(b)

Find : \[\int \frac{\sqrt{\cot x}}{\sin x \cos x} d x \]

(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$$