Question
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{1}{1+\sqrt{\tan\text{x}}}\text{ dx}$
$\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{1}{1+\sqrt{\tan\text{x}}}\text{ dx}$
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Answer
Let $\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{1}{1+\sqrt{\tan\text{x}}}\text{ dx}\ ....(\text{i})$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{1}{1+\sqrt{\tan\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{1}{1+\sqrt{\cot\text{x}}}\text{ dx}\ ....(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\Big(\frac{1}{1+\sqrt{\tan\text{x}}}+\frac{1}{1+\sqrt{\cot\text{x}}}\Big)\text{dx}$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\bigg(\frac{1+\sqrt{\cot\text{x}}+1+\sqrt{\tan\text{x}}}{1+\sqrt{\cot\text{x}}+1+\sqrt{\tan\text{x}}+\sqrt{\tan\text{x}\cot\text{x}}}\bigg)\text{dx}$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{2+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}{2+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\text{dx}=\Big[\text{x}\Big]^{\frac{\pi}{3}}_\frac{\pi}{6}$
$=\frac{\pi}{3}-\frac{\pi}{6}$
$\therefore\ 2\text{I}=\frac{\pi}{6}$
Hence, $\text{I}=\frac{\pi}{12}$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{1}{1+\sqrt{\tan\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{1}{1+\sqrt{\cot\text{x}}}\text{ dx}\ ....(\text{ii})$
Adding (i) and (ii)
$2\text{I}=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\Big(\frac{1}{1+\sqrt{\tan\text{x}}}+\frac{1}{1+\sqrt{\cot\text{x}}}\Big)\text{dx}$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\bigg(\frac{1+\sqrt{\cot\text{x}}+1+\sqrt{\tan\text{x}}}{1+\sqrt{\cot\text{x}}+1+\sqrt{\tan\text{x}}+\sqrt{\tan\text{x}\cot\text{x}}}\bigg)\text{dx}$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\frac{2+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}{2+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}\text{ dx}$
$=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{6}}\text{dx}=\Big[\text{x}\Big]^{\frac{\pi}{3}}_\frac{\pi}{6}$
$=\frac{\pi}{3}-\frac{\pi}{6}$
$\therefore\ 2\text{I}=\frac{\pi}{6}$
Hence, $\text{I}=\frac{\pi}{12}$
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