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: Understanding the Concept:
The limits \(\pi/6\) and \(\pi/3\) add up to \(\pi/2\). So we can use \(\int_a^b f(x)dx = \int_a^b f(a + b - x)dx\) with \(a + b = \pi/2\).

Step 2: Apply the reflection:
Let \(f(x) = \dfrac{\sin x - \cos x}{1 + \sin x\cos x}\). Then
\[ f\left(\frac\pi2 - x\right) = \frac{\cos x - \sin x}{1 + \cos x\sin x} = -f(x) \]

Step 3: Conclude:
So \(I = \int f(x)dx = \int f\left(\tfrac\pi2 - x\right)dx = -I\), giving \(2I = 0\) and \(I = 0\).

Step 4: Check:
The function is negative for \(x < \pi/4\) and positive for \(x > \pi/4\), and the two areas are equal in size. So the integral is 0, option (A).

Final Answer:
The integrand is odd about pi/4, so the integral is 0. \[ \boxed{\text{(A) }0} \]