Definite Integral of Composite Symmetric Function with Cosine

To find the value of the integral $I = \int_{-\pi/2}^{\pi/2} \left( x^2 + \ln \frac{\pi + x}{\pi - x} \right) \cos x \, dx$

we can split the integrand into two separate terms: $I = \int_{-\pi/2}^{\pi/2} x^2 \cos x \, dx + \int_{-\pi/2}^{\pi/2} \ln \left( \frac{\pi + x}{\pi - x} \right) \cos x \, dx$

Let us define: $I_1 = \int_{-\pi/2}^{\pi/2} x^2 \cos x \, dx$ $I_2 = \int_{-\pi/2}^{\pi/2} \ln \left( \frac{\pi + x}{\pi - x} \right) \cos x \, dx$

Step 1: Evaluate $I_2$ Let $f(x) = \ln \left( \frac{\pi + x}{\pi - x} \right) \cos x$. Evaluating $f(-x)$: $f(-x) = \ln \left( \frac{\pi - x}{\pi + x} \right) \cos(-x) = \ln \left( \frac{\pi + x}{\pi - x} \right)^{-1} \cos x = -\ln \left( \frac{\pi + x}{\pi - x} \right) \cos x = -f(x)$

Since $f(x)$ is an odd function and the limits of integration are symmetric about zero, the value of the integral is zero: $I_2 = 0$

Step 2: Evaluate $I_1$ Let $g(x) = x^2 \cos x$. Evaluating $g(-x)$: $g(-x) = (-x)^2 \cos(-x) = x^2 \cos x = g(x)$

Since $g(x)$ is an even function, we can rewrite $I_1$ as: $I_1 = 2 \int_{0}^{\pi/2} x^2 \cos x \, dx$

Using integration by parts ($\int u \, dv = uv - \int v \, du$): Let $u = x^2 \implies du = 2x \, dx$ and $dv = \cos x \, dx \implies v = \sin x$.

$\int x^2 \cos x \, dx = x^2 \sin x - 2 \int x \sin x \, dx$

Applying integration by parts again to $\int x \sin x \, dx$: Let $u = x \implies du = dx$ and $dv = \sin x \, dx \implies v = -\cos x$.

$\int x \sin x \, dx = -x \cos x - \int (-\cos x) \, dx = -x \cos x + \sin x$

Substituting this back into the expression for the indefinite integral: $\int x^2 \cos x \, dx = x^2 \sin x - 2(-x \cos x + \sin x) = x^2 \sin x + 2x \cos x - 2\sin x$

Now, applying the limits from $0$ to $\frac{\pi}{2}$: $\int_{0}^{\pi/2} x^2 \cos x \, dx = \left[ x^2 \sin x + 2x \cos x - 2\sin x \right]_{0}^{\pi/2}$

Evaluating at the upper limit $x = \frac{\pi}{2}$: $\left( \frac{\pi}{2} \right)^2 \sin\left(\frac{\pi}{2}\right) + 2\left(\frac{\pi}{2}\right) \cos\left(\frac{\pi}{2}\right) - 2\sin\left(\frac{\pi}{2}\right) = \frac{\pi^2}{4}(1) + \pi(0) - 2(1) = \frac{\pi^2}{4} - 2$

Evaluating at the lower limit $x = 0$: $0^2 \sin(0) + 2(0)\cos(0) - 2\sin(0) = 0$

Thus, $I_1 = 2 \left( \frac{\pi^2}{4} - 2 \right) = \frac{\pi^2}{2} - 4$

Step 3: Calculate the total integral $I$ $I = I_1 + I_2 = \left( \frac{\pi^2}{2} - 4 \right) + 0 = \frac{\pi^2}{2} - 4$

Hence, the correct option is B.