Question

\(\int _{\pi /6}^{\pi /3}\sin 2x\log (\tan x)\ dx=\ldots\)

• A. \(0\)
• B. \(\frac{\sqrt{3}}{2}-1\)
• C. \(\frac{\sqrt{3}}{2}+\frac{1}{2}\)
• D. \(\left(\frac{\sqrt{3}-1}{2}\right)-\sqrt{3}\)

Hint:

Apply the property that replaces x by (a + b - x); here a + b = pi/2, and the integral turns into its own negative.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Use \(\int _{a}^{b}f(x)dx=\int _{a}^{b}f(a+b-x)dx\). Here \(a+b=\frac{\pi }{2}\).

Step 2: Replace x
\(\sin 2\left(\frac{\pi }{2}-x\right)=\sin 2x\) and \(\log \tan \left(\frac{\pi }{2}-x\right)=\log \cot x=-\log \tan x\).

Step 3: Form an equation
So \(I=-I\), hence \(2I=0\).

Step 4: Conclude
\(I=0\).

Final Answer:
The value of the integral is 0. \[ \boxed{\text{(A) }0} \]