$\text{I}=\int\Big(\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}\Big)\text{dx}=\int\frac{\cos\text{x}+\sin\text{x}}{\sqrt{\sin\text{x}\cos\text{x}}}\text{dx}$ Putting sin x – cos x = t, so that (cos x + sin x) dx = dt and sin x cos x =$\frac1 2$(1 – t2) $\therefore\ \text{I}=\sqrt2\int\frac{\text{dt}}{1-\text{t}^2}=\sqrt2\sin^{-1}\text{t}+\text{c}$ $=\sqrt2\sin^{-1}(\sin\text{x}-\cos\text{x})+\text{c}$
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