Question

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

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

Hint:

For integrals of the form $\int_{0}^{\pi} x f(\sin x) \, dx$, the $x$ can always be replaced by $\frac{\pi}{2}$ outside the integral. This problem simplifies directly to $\frac{\pi}{2} \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} \, dx$.

Answer 1

The Correct Option is C

Concept: To solve this integral, we apply the definite integral property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$. This technique is effective for removing the $x$ factor from the numerator.

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

Step 2: Combine the integrals.
Adding the two expressions for $I$: \[ 2I = \int_{0}^{\pi} \frac{(x + \pi - x) \sin x}{\sin^2 x + 2 \cos^2 x} \, dx = \int_{0}^{\pi} \frac{\pi \sin x}{\sin^2 x + 2 \cos^2 x} \, dx \] Using $\sin^2 x = 1 - \cos^2 x$: \[ 2I = \pi \int_{0}^{\pi} \frac{\sin x}{1 - \cos^2 x + 2 \cos^2 x} \, dx = \pi \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} \, dx \] \[ I = \frac{\pi}{2} \int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} \, dx \]

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