Question

The value of the integral $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{32 \cos^4 x}{1 + e^{\sin x}} dx$ is:

• A. $4\pi + 2$
• B. $3\pi + 8$
• C. $3\pi + 4$
• D. $4\pi + 3$

Hint:

Use the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$ to simplify the denominator. Then apply the power reduction formula for $\cos^4 x$.

Answer 1

The Correct Option is B

Another way to solve $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{32 \cos^4 x}{1 + e^{\sin x}} dx$ is by splitting the integral into its even and odd components using the property $\int_{-a}^a f(x) dx = \int_0^a [f(x) + f(-x)] dx$.
Let $f(x) = \frac{32 \cos^4 x}{1 + e^{\sin x}}$. Then:
$$f(x) + f(-x) = \frac{32 \cos^4 x}{1 + e^{\sin x}} + \frac{32 \cos^4 (-x)}{1 + e^{\sin (-x)}}$$
$$= 32 \cos^4 x \left( \frac{1}{1 + e^{\sin x}} + \frac{1}{1 + e^{-\sin x}} \right) = 32 \cos^4 x \left( \frac{1}{1 + e^{\sin x}} + \frac{e^{\sin x}}{e^{\sin x} + 1} \right)$$
$$= 32 \cos^4 x \left( \frac{1 + e^{\sin x}}{1 + e^{\sin x}} \right) = 32 \cos^4 x$$
Thus, the integral is reduced to:
$$I = \int_{0}^{\frac{\pi}{4}} 32 \cos^4 x dx$$
Using the power-reduction formula $\cos^4 x = \frac{3 + 4 \cos(2x) + \cos(4x)}{8}$, we have:
$$I = 32 \int_{0}^{\frac{\pi}{4}} \frac{3 + 4 \cos(2x) + \cos(4x)}{8} dx = 4 \int_{0}^{\frac{\pi}{4}} (3 + 4 \cos(2x) + \cos(4x)) dx$$
Integrating term by term:
$$I = 4 \left[ 3x + 2 \sin(2x) + \frac{1}{4} \sin(4x) \right]_0^{\frac{\pi}{4}}$$
Substituting the upper limit $\frac{\pi}{4}$:
$$I = 4 \left( \frac{3\pi}{4} + 2 \sin(\frac{\pi}{2}) + \frac{1}{4} \sin(\pi) \right) = 4 \left( \frac{3\pi}{4} + 2(1) + 0 \right)$$
$$I = 3\pi + 8$$
The final result is $3\pi + 8$.