Question

\(\int _{π/6}^{π/3}\frac{sinx-cosx}{1+sinxcosx}\,dx =\)

• A. \(0\)
• B. \(\frac{π}{12}\)
• C. \(\frac{π}{24}\)
• D. \(\frac{π^2}{2}\)

Hint:

Use the reflection x -> pi/2 - x on a symmetric interval.

Answer 1

The Correct Option is A

Step 1: Symmetry about pi/4:
The interval $[\pi/6, \pi/3]$ is centred at $\pi/4$. Put $x = \pi/4 + u$ with $u\in[-\pi/12, \pi/12]$.

Step 2: Transform:
$\sin x - \cos x = \sqrt2\sin u$ and $\sin x\cos x = \tfrac12\sin2x = \tfrac12\cos2u$, which is even in $u$.

Step 3: Result:
The integrand is $\dfrac{\sqrt2\sin u}{1 + \tfrac12\cos 2u}$, an odd function of $u$ on a symmetric interval, so the integral is $0$, option (A).

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