Question

If $ a = 2n $ and $ b = 2m+1 $ for all $ m, n \in \mathbb{N} $, then \[ \int_{-\pi}^{\pi} e^{\sin^2 x} \cot^{b}(2n+1) x \, dx = \]

• A. \( 0 \)
• B. \( 1 \)
• C. \( -1 \)
• D. \( \pi \)

Hint:

If the integrand is an odd function over a symmetric interval around zero (like \([-\pi, \pi]\)), the integral evaluates to zero.

Answer 1

The Correct Option is A

Step 1: Analyze the integrand.
We are given:
\( a = 2n \), an even number,
\( b = 2m+1 \), an odd number,
So the integrand becomes: \( f(x) = e^{\sin^2 x} \cdot \cot^b((2n+1)x) \)
Step 2: Check the symmetry of the function.
We check whether the function is odd over the symmetric interval \([-\pi, \pi]\): \[ f(-x) = e^{\sin^2 (-x)} \cdot \cot^b((2n+1)(-x)) = e^{\sin^2 x} \cdot [-\cot((2n+1)x)]^b \] Since \( b \) is odd, this becomes: \[ f(-x) = -e^{\sin^2 x} \cdot \cot^b((2n+1)x) = -f(x) \] So, \( f(x) \) is an odd function.
Step 3: Use the property of definite integrals.
If \( f(x) \) is an odd function, then: \[ \int_{-a}^{a} f(x)\, dx = 0 \] \[ \Rightarrow \int_{-\pi}^{\pi} e^{\sin^2 x} \cot^b((2n+1)x) \, dx = 0 \]