Question

Prove that \[ \int_{0}^{\pi} \frac{x\sin x}{1+\cos^2 x}\, dx = \frac{\pi^2}{4}. \]

Hint:

For symmetric integrals involving $f(x)$ and $f(\pi-x)$, always try the property $I=\int_0^\pi f(x)dx = \int_0^\pi f(\pi-x)dx$.

Answer 1

Step 1: Let \[ I = \int_{0}^{\pi} \frac{x\sin x}{1+\cos^2 x}\, dx. \]

Step 2: Use property of definite integrals.
\[ I = \int_{0}^{\pi} f(x)\, dx = \int_{0}^{\pi} f(\pi-x)\, dx. \] Here, \[ f(x) = \frac{x\sin x}{1+\cos^2 x}. \] So, \[ f(\pi-x) = \frac{(\pi-x)\sin(\pi-x)}{1+\cos^2(\pi-x)}. \] Since $\sin(\pi-x)=\sin x$, $\cos(\pi-x)=-\cos x$ and $\cos^2(\pi-x)=\cos^2 x$: \[ f(\pi-x) = \frac{(\pi-x)\sin x}{1+\cos^2 x}. \]

Step 3: Add $I+I$.
\[ 2I = \int_{0}^{\pi} \left[ \frac{x\sin x}{1+\cos^2 x} + \frac{(\pi-x)\sin x}{1+\cos^2 x} \right] dx. \] \[ 2I = \int_{0}^{\pi} \frac{\pi\sin x}{1+\cos^2 x} \, dx. \] \[ I = \frac{\pi}{2} \int_{0}^{\pi} \frac{\sin x}{1+\cos^2 x} \, dx. \]

Step 4: Substitute $u=\cos x$.
\[ du = -\sin x \, dx. \] When $x=0 $\Rightarrow$ u=1$, when $x=\pi $\Rightarrow$ u=-1$. So, \[ I = \frac{\pi}{2}\int_{1}^{-1} \frac{-du}{1+u^2}. \] \[ I = \frac{\pi}{2}\int_{-1}^{1} \frac{du}{1+u^2}. \] \[ I = \frac{\pi}{2}\left[\tan^{-1}u\right]_{-1}^{1}. \] \[ I = \frac{\pi}{2}\left(\tan^{-1}(1) - \tan^{-1}(-1)\right). \] \[ I = \frac{\pi}{2}\left(\frac{\pi}{4} - \left(-\frac{\pi}{4}\right)\right). \] \[ I = \frac{\pi}{2}\cdot \frac{\pi}{2} = \frac{\pi^2}{4}. \]

Final Answer: \[ \boxed{\frac{\pi^2}{4}} \]