Evaluate: _ /6 ^ /3 dx 1 + . - Vidyadip

Question

Evaluate: $\int\limits_{\pi/6}^{\pi/3}\frac{\text{dx}}{1 + \sqrt{\cot\text{x}}}.$

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Answer

$\text{I} = \int\limits_{\pi/6}^{\pi/3}\frac{\text{dx}}{1 + \sqrt{\cot\text{x}}} = \int\limits_{\pi/6}^{\pi/3}\frac{\sqrt{\sin\text{x}}}{\sqrt{\sin\text{x}} + \sqrt{\cos\text{x}}}\text{dx}$
$ = \int\limits_{\pi/6}^{\pi/3}\frac{\sqrt{\sin\bigg(\frac{\pi}{3} + \frac{\pi}{6} - \text{x}}\bigg)}{\sqrt{\sin\bigg(\frac{\pi}{3} + \frac{\pi}{6} - \text{x}\bigg) + \sqrt{\cos\bigg(\frac{\pi}{3} + \frac{\pi}{6} - \text{x}}\bigg)}}\text{dx}$
$\therefore\text{I} = \int\limits_{\pi/6}^{\pi/3}\frac{\sqrt{\cos\text{x}}}{\sqrt{\cos\text{x}} +\sqrt{\sin\text{x}}}\text{dx}$
Adding we get, $2 \text{I} = \int\limits_{\pi/6}^{\pi/3}\text{dx} = \big[\text{x}\big]_{\pi/6}^{\pi/3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$
$\therefore\text{I} = \frac{\pi}{12}.$

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