Calculate Definite Integral of Trigonometric Function

To find the value of the definite integral $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{32 \cos^4 x}{1 + e^{\sin x}} dx$

we can use the fundamental property of definite integrals: $\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx$

Here, $a = -\frac{\pi}{4}$ and $b = \frac{\pi}{4}$, so $a + b - x = -x$. Replacing $x$ with $-x$ in the integrand gives: $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{32 \cos^4(-x)}{1 + e^{\sin(-x)}} dx$

Since $\cos(-x) = \cos x$ and $\sin(-x) = -\sin x$, this simplifies to: $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{32 \cos^4 x}{1 + e^{-\sin x}} dx$

Multiplying the numerator and denominator by $e^{\sin x}$, we get: $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{32 \cos^4 x \cdot e^{\sin x}}{e^{\sin x} + 1} dx$

Now, adding the two representations of $I$: $2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \left( \frac{32 \cos^4 x}{1 + e^{\sin x}} + \frac{32 \cos^4 x \cdot e^{\sin x}}{1 + e^{\sin x}} \right) dx$ $2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 32 \cos^4 x \left( \frac{1 + e^{\sin x}}{1 + e^{\sin x}} \right) dx$ $2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 32 \cos^4 x \, dx$

Since $\cos^4 x$ is an even function ($f(-x) = f(x)$), we can write: $2I = 2 \int_{0}^{\frac{\pi}{4}} 32 \cos^4 x \, dx$ $I = 32 \int_{0}^{\frac{\pi}{4}} \cos^4 x \, dx$

To integrate $\cos^4 x$, we use the double-angle identity $\cos^2 x = \frac{1 + \cos 2x}{2}$: $\cos^4 x = \left( \frac{1 + \cos 2x}{2} \right)^2 = \frac{1 + 2\cos 2x + \cos^2 2x}{4}$

Using $\cos^2 2x = \frac{1 + \cos 4x}{2}$: $\cos^4 x = \frac{1}{4} \left( 1 + 2\cos 2x + \frac{1 + \cos 4x}{2} \right) = \frac{1}{8} \left( 3 + 4\cos 2x + \cos 4x \right)$

Substituting this back into the integral: $I = 32 \int_{0}^{\frac{\pi}{4}} \frac{1}{8} \left( 3 + 4\cos 2x + \cos 4x \right) dx$ $I = 4 \int_{0}^{\frac{\pi}{4}} \left( 3 + 4\cos 2x + \cos 4x \right) dx$

Integrating term by term: $I = 4 \left[ 3x + 2\sin 2x + \frac{\sin 4x}{4} \right]_{0}^{\frac{\pi}{4}}$

Evaluating at the upper limit $x = \frac{\pi}{4}$: $I = 4 \left[ 3\left(\frac{\pi}{4}\right) + 2\sin\left(\frac{\pi}{2}\right) + \frac{\sin \pi}{4} \right]$ $I = 4 \left[ \frac{3\pi}{4} + 2(1) + 0 \right]$ $I = 4 \left( \frac{3\pi}{4} + 2 \right) = 3\pi + 8$

Thus, the correct option is B.