Evaluate ∫₀^π (x sin x)/(1 + cos² x) , dx — JEE Mathematics

Let $I = \int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x} \, dx$. --- (1)

Apply the property $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$:

$$I = \int_{0}^{\pi} \frac{(\pi - x) \sin(\pi - x)}{1 + \cos^2(\pi - x)} \, dx$$

Since $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x \implies \cos^2(\pi - x) = \cos^2 x$:

$$I = \int_{0}^{\pi} \frac{(\pi - x) \sin x}{1 + \cos^2 x} \, dx \quad \dots (2)$$

Add equations (1) and (2):

$$2I = \int_{0}^{\pi} \frac{(x + \pi - x) \sin x}{1 + \cos^2 x} \, dx = \pi \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} \, dx$$

$$I = \frac{\pi}{2} \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} \, dx$$

Let $u = \cos x$, then $du = -\sin x \, dx$.
When $x = 0$, $u = 1$. When $x = \pi$, $u = -1$.

$$I = \frac{\pi}{2} \int_{1}^{-1} \frac{-du}{1 + u^2} = \frac{\pi}{2} \int_{-1}^{1} \frac{du}{1 + u^2}$$

$$I = \frac{\pi}{2} \left[ \tan^{-1} u \right]_{-1}^{1} = \frac{\pi}{2} \left( \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) \right) = \frac{\pi}{2} \left( \frac{\pi}{2} \right) = \frac{\pi^2}{4}$$

Answer: $\frac{\pi^2}{4}$