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: Analyze Odd/Even Functions
$f(x) = x^2$ is an even function. $g(x) = \log(\frac{\pi-x}{\pi+x}) \cos x$ is an odd function because $\log(1/A) = -\log A$.
Step 2: Property of Integration
$\int_{-a}^a g(x) dx = 0$ for odd functions. The integral reduces to $\int_{-\pi/2}^{\pi/2} x^2 dx$.
Step 3: Calculation
$2 \int_0^{\pi/2} x^2 dx = 2 [x^3/3]_0^{\pi/2}$. $= 2 \cdot \frac{1}{3} \cdot \frac{\pi^3}{8} = \frac{\pi^3}{12}$.
Step 4: Conclusion
The value is $\pi^3/12$.
Final Answer:(B)