Question

Evaluate: \[ \int_{0}^{\pi} \frac{x\sin x}{1+\cos^2 x}\,dx \]

• A. $\dfrac{\pi^2}{4}$
• B. $\dfrac{\pi^2}{2}$
• C. $\dfrac{\pi^2}{3}$
• D. $\pi^2$

Hint:

For integrals of the form $\int_0^\pi x f(\sin x,\cos x)\,dx$, use the property $f(x)+f(\pi-x)$ to simplify the calculation.

Answer 1

The Correct Option is A

Step 1: Let \[ I=\int_{0}^{\pi} \frac{x\sin x}{1+\cos^2 x}\,dx \]
Step 2: Replace $x$ by $\pi-x$ in the integral: \[ I=\int_{0}^{\pi} \frac{(\pi-x)\sin x}{1+\cos^2 x}\,dx \]
Step 3: Add the two expressions for $I$: \[ 2I=\int_{0}^{\pi} \frac{\pi\sin x}{1+\cos^2 x}\,dx \]
Step 4: Take $\pi$ outside the integral: \[ 2I=\pi\int_{0}^{\pi} \frac{\sin x}{1+\cos^2 x}\,dx \]
Step 5: Substitute $u=\cos x$, so $du=-\sin x\,dx$. When $x=0$, $u=1$; when $x=\pi$, $u=-1$. \[ \int_{0}^{\pi} \frac{\sin x}{1+\cos^2 x}\,dx =\int_{-1}^{1} \frac{du}{1+u^2} =\left[\tan^{-1}u\right]_{-1}^{1} =\frac{\pi}{2} \]
Step 6: Hence, \[ 2I=\pi\cdot\frac{\pi}{2}=\frac{\pi^2}{2} \Rightarrow I=\frac{\pi^2}{4} \]