Definite Integrals — Maths STD 12 Science — Question

3. $\frac{\pi}{12}$

Solution:

Let, $\text{I}=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\frac{1}{1+\sqrt{\cot\text{x}}}\text{dx}\ ...{\text{(i)}}$

$=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\frac{1}{\sqrt{\cot\big(\frac{\pi}{3}+\frac{\pi}{6}-\text{x}\big)}}\text{dx}$ $\bigg[\text{using}\int\limits^\text{b}_\text{a}\text{f}(\text{x})\text{dx}=\int\limits^\text{b}_\text{a}\text{f}\big(\text{a}+\text{b}-\text{x}\big)\text{dx}\bigg]$

$=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\frac{1}{1+\sqrt{\tan\text{x}}}\text{dx}\ ...(\text{ii})$

Adding (i) and (ii) we get

$2\text{I}=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\bigg[\frac{1}{1+\sqrt{\cot\text{x}}}+\frac{1}{1+\sqrt{\tan\text{x}}}\bigg]\text{dx}$

$=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\frac{2+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}{\big(1+\sqrt{\cot\text{x}}\big)+\big(1+\sqrt{\tan\text{x}}\big)}\text{ dx}$

$=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\Bigg[\frac{2+\sqrt{\cot\text{x}}+\sqrt{\tan\text{x}}}{2+\sqrt{\cot\text{x}+\sqrt{\tan\text{x}}}}\Bigg]\text{dx}$

$=\int\limits^\frac{\pi}{3}_\frac{\pi}{6}\text{dx}$

$=\big[\text{x}\big]^\frac{\pi}{3}_\frac{\pi}{6}$

$=\frac{\pi}{3}-\frac{\pi}{6}$

$=\frac{\pi}{6}$

Hence, $\text{I}=\frac{\pi}{12}$