Question

Evaluate \( \displaystyle \int_{-\pi}^{\pi} \frac{2x}{1+\cos^2 x}\,dx \)

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

Hint:

Always check whether the integrand is odd or even before evaluating definite integrals over symmetric limits.

Answer 1

The Correct Option is B

Step 1: Identify the nature of the integrand.
The numerator \( 2x \) is an odd function, while the denominator \( 1+\cos^2 x \) is an even function.
Step 2: Determine the parity of the integrand.
An odd function divided by an even function remains an odd function. Hence, \[ f(x) = \frac{2x}{1+\cos^2 x} \] is an odd function.
Step 3: Use the property of definite integrals.
For any odd function \( f(x) \), \[ \int_{-a}^{a} f(x)\,dx = 0 \] Step 4: Apply the property.
\[ \int_{-\pi}^{\pi} \frac{2x}{1+\cos^2 x}\,dx = 0 \] Step 5: Conclusion.
Hence, the value of the given integral is \( 0 \).