\(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\displaystyle\frac{dx}{1+\sqrt{\tan x}}\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{\pi}{12}\)
Let I be the integral. By the property \(\int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(a + b - x)\,dx\), with \(a + b = \frac{\pi}{2}\), replace x by \(\frac{\pi}{2} - x\); then \(\tan x\) becomes \(\cot x\):
\(\displaystyle I = \int_{\pi/6}^{\pi/3} \frac{dx}{1 + \sqrt{\cot x}}\)
Multiplying the numerator and denominator by \(\sqrt{\tan x}\):
\(\displaystyle I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\tan x}}{\sqrt{\tan x} + 1}\,dx\)
Adding the two forms of I:
\(\displaystyle 2I = \int_{\pi/6}^{\pi/3} \frac{1 + \sqrt{\tan x}}{1 + \sqrt{\tan x}}\,dx = \int_{\pi/6}^{\pi/3} 1\,dx\)
\(\displaystyle 2I = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}\)
\(\displaystyle I = \frac{\pi}{12}\)
Shortcut: The integrand equals \(\frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}}\), which has the form \(\frac{f(x)}{f(x) + f(a + b - x)}\). Every such integral from a to b equals half the length of the interval, \(\frac{b - a}{2} = \frac{1}{2}\left(\frac{\pi}{3} - \frac{\pi}{6}\right) = \frac{\pi}{12}\).
NCERT (2026-27 edition):
Some Properties of Definite Integrals
