Question

$\int_0^\pi \frac{x \sin x}{1 + \cos^2 x} \, dx =$
Identify the correct option from the following:

• A. $\frac{\pi^2}{4}$
• B. $\frac{\pi}{2}$
• C. $\frac{\pi^2}{2}$
• D. $\frac{\pi}{4}$

Hint:

For integrals of the form $\int_0^\pi x f(\sin x, \cos x) \, dx$, use the substitution $u = \pi - x$ to simplify and exploit symmetry.

Answer 1

The Correct Option is A

Step 1: Use symmetry of the integral
Consider $I = \int_0^\pi \frac{x \sin x}{1 + \cos^2 x} \, dx$. Substitute $u = \pi - x$, $du = -dx$, $\sin u = \sin x$, $\cos u = -\cos x$, so $I = \int_0^\pi \frac{(\pi - u) \sin u}{1 + \cos^2 u} \, du$. Add: $2I = \int_0^\pi \frac{(x + (\pi - x)) \sin x}{1 + \cos^2 x} \, dx = \pi \int_0^\pi \frac{\sin x}{1 + \cos^2 x} \, dx$. Step 2: Compute the simplified integral
$I = \frac{\pi}{2} \int_0^\pi \frac{\sin x}{1 + \cos^2 x} \, dx$. Let $v = \cos x$, $dv = -\sin x \, dx$, limits $x = 0$ to $\pi$, $v = 1$ to $-1$. Then $\int_0^\pi \frac{\sin x}{1 + \cos^2 x} \, dx = \int_1^{-1} \frac{-dv}{1 + v^2} = \int_{-1}^1 \frac{1}{1 + v^2} \, dv = \tan^{-1} v \big|_{-1}^1 = \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) = \pi$. So $I = \frac{\pi}{2} \cdot \pi = \frac{\pi^2}{4}$. Step 3: Match with options
The result $\frac{\pi^2}{4}$ matches option (1).