\(\int_{\pi/6}^{\pi/3}\displaystyle\frac{\tan x}{\tan x + \cot x}\,dx\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{\pi}{12}\)
The integrand simplifies to \(\sin^2 x\):
\(\displaystyle \tan x + \cot x = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x}\)
\(\displaystyle \frac{\tan x}{\tan x + \cot x} = \frac{\sin x}{\cos x} \cdot \sin x \cos x = \sin^2 x\)
Using \(\sin^2 x = \frac{1 - \cos 2x}{2}\):
\(\displaystyle I = \int_{\pi/6}^{\pi/3} \frac{1 - \cos 2x}{2}\,dx = \left[\frac{x}{2} - \frac{\sin 2x}{4}\right]_{\pi/6}^{\pi/3}\)
\(\displaystyle = \left(\frac{\pi}{6} - \frac{1}{4}\sin\frac{2\pi}{3}\right) - \left(\frac{\pi}{12} - \frac{1}{4}\sin\frac{\pi}{3}\right)\)
Since \(\sin\frac{2\pi}{3} = \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}\), the sine terms cancel:
\(\displaystyle I = \frac{\pi}{6} - \frac{\pi}{12} = \frac{\pi}{12}\)
Shortcut: By the property \(\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx\) with \(a + b = \frac{\pi}{2}\), x becomes \(\frac{\pi}{2} - x\), which swaps tan and cot:
\(\displaystyle I = \int_{\pi/6}^{\pi/3} \frac{\cot x}{\cot x + \tan x}\,dx\)
\(\displaystyle 2I = \int_{\pi/6}^{\pi/3} 1\,dx = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}\)
\(\displaystyle I = \frac{\pi}{12}\)