Question

The value of the integral \(\int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x} \, dx\) is:

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

Hint:

For any integral of the form $\int_{0}^{\pi} x f(\sin x) \, dx$, you can use the shortcut: \[ \int_{0}^{\pi} x f(\sin x) \, dx = \frac{\pi}{2} \int_{0}^{\pi} f(\sin x) \, dx \] This immediately simplifies the problem by removing the $x$ multiplier.

Answer 1

The Correct Option is A

Concept: To solve this integral, we use the definite integral property
$\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$.
This allows us to eliminate the $x$ term in the numerator, which is a common technique for integrands involving $x \sin x$ or $x \cos x$.

Step 1: Apply the integral property.
Let $I = \int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x} \, dx$.
Replacing $x$ with $(\pi - x)$: \[ I = \int_{0}^{\pi} \frac{(\pi - x) \sin(\pi - x)}{1 + \cos^2(\pi - x)} \, dx \] Since $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$ (so $\cos^2(\pi - x) = \cos^2 x$): \[ I = \int_{0}^{\pi} \frac{(\pi - x) \sin x}{1 + \cos^2 x} \, dx \]

Step 2: Sum the two forms of the integral.
Adding the original $I$ and the modified $I$: \[ 2I = \int_{0}^{\pi} \frac{x \sin x + (\pi - x) \sin x}{1 + \cos^2 x} \, dx \] \[ 2I = \int_{0}^{\pi} \frac{\pi \sin x}{1 + \cos^2 x} \, dx \implies I = \frac{\pi}{2} \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} \, dx \]

Step 3: Integrate using substitution.
Let $u = \cos x$, then $du = -\sin x \, dx$. When $x = 0, u = 1$; when $x = \pi, u = -1$. \[ I = \frac{\pi}{2} \int_{1}^{-1} \frac{-du}{1 + u^2} = \frac{\pi}{2} \int_{-1}^{1} \frac{du}{1 + u^2} \] \[ I = \frac{\pi}{2} \left[ \tan^{-1} u \right]_{-1}^{1} = \frac{\pi}{2} \left( \frac{\pi}{4} - \left( -\frac{\pi}{4} \right) \right) \] \[ I = \frac{\pi}{2} \left( \frac{\pi}{2} \right) = \frac{\pi^2}{4} \]