Question

\( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^2 + \log (\frac{\pi - x}{\pi + x}) \cdot \cos x) dx = \)}

• A. 0
• B. \( \frac{\pi^3}{12} \)
• C. \( \frac{\pi^2}{2} - 4 \)
• D. \( \frac{\pi^2}{2} + 4 \)

Hint:

$\log(\frac{a-x}{a+x})$ is a classic odd function. Multiplying it by an even function ($\cos x$) keeps it odd.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Use the properties of integration for odd and even functions over symmetric intervals \( [-a, a] \).
Step 2: Key Formula or Approach:
1. \( \int_{-a}^{a} f(x) dx = 2 \int_{0}^{a} f(x) dx \) if \( f \) is even.
2. \( \int_{-a}^{a} f(x) dx = 0 \) if \( f \) is odd.
Step 3: Detailed Explanation:
Let \( f(x) = x^2 \cos x \) and \( g(x) = \log(\frac{\pi - x}{\pi + x}) \cos x \).
\( f(-x) = (-x)^2 \cos(-x) = x^2 \cos x \to \) Even.
\( g(-x) = \log(\frac{\pi + x}{\pi - x}) \cos(-x) = -\log(\frac{\pi - x}{\pi + x}) \cos x \to \) Odd.
Thus, the integral becomes \( 2 \int_{0}^{\pi/2} x^2 \cos x dx + 0 \).
Use integration by parts twice for \( \int x^2 \cos x dx \):
\[ \int x^2 \cos x dx = x^2 \sin x - \int 2x \sin x dx \] \[ = x^2 \sin x - [2x(-\cos x) - \int 2(-\cos x) dx] \] \[ = x^2 \sin x + 2x \cos x - 2\sin x \] Evaluate from 0 to \( \pi/2 \):
\[ 2 \left[ (\frac{\pi^2}{4})(1) + 0 - 2(1) \right] - 0 = 2(\frac{\pi^2}{4} - 2) = \frac{\pi^2}{2} - 4 \] Step 4: Final Answer:
The result is \( \frac{\pi^2}{2} - 4 \).