Question
Evaluate the following integrals:$\int\limits^{\frac{\pi}{2}}_0\frac{\sqrt{\cot\text{x}}}{\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}\text{ dx}$

Answer

Let $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{\sqrt{\cot\text{x}}}{\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}\text{ dx}\ ...(\text{i})$$=\int\limits^{\frac{\pi}{2}}_0\frac{\sqrt{\cot\big(\frac{\pi}{2}-\text{x}\big)}}{\sqrt{\cot\big(\frac{\pi}{2}-\text{x}\big)}+\sqrt{\tan\big(\frac{\pi}{2}-\text{x}\big)}}\text{ dx}$ $\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$
$=\int\limits^{\frac{\pi}{2}}_0\frac{\sqrt{\tan\text{x}}}{\sqrt{\tan\text{x}}+\sqrt{\cot\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii)$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\bigg(\frac{\sqrt{\cot\text{x}}}{\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}+\frac{\sqrt{\tan\text{x}}}{\sqrt{\tan\text{x}}+\sqrt{\cot\text{x}}}\bigg)\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\text{dx}$
$=\Big[\text{x}\Big]^{\frac{\pi}{2}}_0=\frac{\pi}{2}$
Hence, $\text{I}=\frac{\pi}{4}$

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