Question
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\sqrt{\tan\text{x}}}\text{ dx}$

Answer

Let $\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\sqrt{\tan\text{x}}}\text{ dx}\ ....(\text{i})$
$=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\sqrt{\tan\big(\frac{\pi}{2}-\text{x}}\big)}\text{dx}$ $\Big[\text{Using},\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\int\limits^{\text{a}}_0\text{f}(\text{a}-\text{x})\text{dx}\Big]$
$=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\sqrt{\cot\text{x}}}\text{ dx}\ ...(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{\frac{\pi}{2}}_0\frac{1}{1+\sqrt{\tan\text{x}}}+\frac{1}{1+\sqrt{\cot\text{x}}}\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{1+\sqrt{\cot\text{x}}+1+\sqrt{\tan\text{x}}}{\big(1+\sqrt{\tan\text{x}}\big)\big(1+\sqrt{\cot\text{x}}\big)}\text{dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{1+\sqrt{\cot\text{x}}+1+\sqrt{\tan\text{x}}}{1+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}+\sqrt{\tan\text{x}\cot\text{x}}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\frac{2+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}{2+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}\text{ dx}$
$=\int\limits^{\frac{\pi}{2}}_0\text{dx}$
$=\Big[\text{x}\Big]^{\frac{\pi}{2}}_0$
$=\frac{\pi}{2}$
Hence, $\text{I}=\frac{\pi}{4}$

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