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

Answer

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

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