Question

Evaluate: \[ \int_0^\pi \frac{x \sin x}{1 + \cos^2 x} \, dx. \]

Hint:

When evaluating trigonometric integrals, try substitution or integration by parts.

Answer 1

Step 1: Substitution.
Let \( u = \cos x \), so that \( du = -\sin x \, dx \). The limits change as follows: - When \( x = 0 \), \( u = \cos(0) = 1 \), - When \( x = \pi \), \( u = \cos(\pi) = -1 \). Thus, the integral becomes: \[ \int_1^{-1} \frac{-x}{1 + u^2} \, du. \] After simplifying, we evaluate this integral. We may also use integration by parts if needed to resolve this fully. The final result involves simplifying trigonometric expressions.
Step 2: Conclusion.
The final result is the evaluated value of the integral.