Question

Find the value of the integral: \[ I = \int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x} dx. \]

Hint:

For definite integrals with trigonometric functions, use substitutions and symmetry properties to simplify the computation.

Answer 1

Using the integration by parts method, let: - \( u = x \Rightarrow du = dx \).
- \( dv = \frac{\sin x}{1 + \cos^2 x} dx \). To integrate \( dv \), substitute \( t = 1 + \cos^2 x \Rightarrow dt = -2 \cos x \sin x dx \). Thus, rewriting the integral: \[ I = \int x d \left( \tan^{-1} (\cos x) \right). \] Using integration by parts: \[ I = x \tan^{-1} (\cos x) \Big|_{0}^{\pi} - \int_{0}^{\pi} \tan^{-1} (\cos x) dx. \] From symmetry properties, the integral simplifies to: \[ I = \frac{\pi}{4} \pi - \frac{\pi^2}{4} = \frac{\pi^2}{4}. \]