Question

The value of $\int_{-\pi}^{\pi} \frac{x\sin x}{1+\cos^2 x}\,dx =$

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

Hint:

A highly valuable standard definite integral formula to memorize for competitive exams is: $$\int_{0}^{\pi} \frac{x\sin x}{1 + \cos^2 x} dx = \frac{\pi^2}{4}$$ Since our even function property showed that the original problem over symmetric bounds is exactly twice this value, we multiply by $2$ to instantly get $\frac{\pi^2}{2}$ without executing any steps!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to evaluate a definite integral containing a product of an algebraic variable and trigonometric functions over a symmetric interval $[-\pi, \pi]$.

Step 2: Key Formula or Approach:
We can apply properties of definite integrals: 1. If $f(x)$ is an even function, $\int_{-a}^{a} f(x) dx = 2\int_{0}^{a} f(x) dx$. 2. King's Property: $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$. We will also use integration by substitution for the simplified version.

Step 3: Detailed Explanation:
Let the integrand be $f(x) = \frac{x\sin x}{1 + \cos^2 x}$. Test for parity: $$f(-x) = \frac{(-x)\sin(-x)}{1 + \cos^2(-x)} = \frac{(-x)(-\sin x)}{1 + \cos^2 x} = \frac{x\sin x}{1 + \cos^2 x} = f(x)$$ Since $f(-x) = f(x)$, the function is even. Therefore, we rewrite the integral: $$I = 2\int_{0}^{\pi} \frac{x\sin x}{1 + \cos^2 x} dx \quad \text{--- (Equation 1)}$$ Apply King's Property ($\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$) to Equation 1: $$I = 2\int_{0}^{\pi} \frac{(\pi - x)\sin(\pi - x)}{1 + \cos^2(\pi - x)} dx$$ Since $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x \Rightarrow \cos^2(\pi - x) = \cos^2 x$: $$I = 2\int_{0}^{\pi} \frac{(\pi - x)\sin x}{1 + \cos^2 x} dx \quad \text{--- (Equation 2)}$$ Add Equation 1 and Equation 2: $$2I = 2\int_{0}^{\pi} \frac{x\sin x + (\pi - x)\sin x}{1 + \cos^2 x} dx$$ $$2I = 2\int_{0}^{\pi} \frac{\pi\sin x}{1 + \cos^2 x} dx \Rightarrow I = \pi\int_{0}^{\pi} \frac{\sin x}{1 + \cos^2 x} dx$$ To integrate this, use substitution: let $t = \cos x$, so $dt = -\sin x dx \Rightarrow \sin x dx = -dt$. Change boundaries: when $x = 0 \Rightarrow t = 1$; when $x = \pi \Rightarrow t = -1$. $$I = \pi\int_{1}^{-1} \frac{-dt}{1 + t^2} = \pi\int_{-1}^{1} \frac{dt}{1 + t^2}$$ Since $\frac{1}{1+t^2}$ is an even function: $$I = 2\pi\int_{0}^{1} \frac{dt}{1 + t^2} = 2\pi \Big[ \tan^{-1} t \Big]_{0}^{1}$$ $$I = 2\pi \left( \tan^{-1}(1) - \tan^{-1}(0) \right) = 2\pi \left( \frac{\pi}{4} - 0 \right) = \frac{2\pi^2}{4} = \frac{\pi^2}{2}$$

Step 4: Final Answer:
The value of the definite integral is $\frac{\pi^2}{2}$, which corresponds to option (A).