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

To solve the given integral $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{32 \cos^4 x}{1 + e^{\sin x}} dx$, we utilize the property of definite integrals:
$$\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$$
In this case, $a = -\pi/4$ and $b = \pi/4$, so $a+b-x = -x$. Applying this transformation:
$$I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{32 \cos^4 (-x)}{1 + e^{\sin (-x)}} dx = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{32 \cos^4 x}{1 + e^{-\sin x}} dx$$
Since $e^{-\sin x} = \frac{1}{e^{\sin x}}$, the integral becomes:
$$I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{32 \cos^4 x \cdot e^{\sin x}}{e^{\sin x} + 1} dx$$
Adding the original and transformed forms of the integral:
$$2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \left( \frac{32 \cos^4 x}{1 + e^{\sin x}} + \frac{32 \cos^4 x \cdot e^{\sin x}}{1 + e^{\sin x}} \right) dx$$
$$2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{32 \cos^4 x (1 + e^{\sin x})}{1 + e^{\sin x}} dx = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 32 \cos^4 x dx$$
Since $32 \cos^4 x$ is an even function, we can simplify:
$$2I = 2 \int_{0}^{\frac{\pi}{4}} 32 \cos^4 x dx \implies I = 32 \int_{0}^{\frac{\pi}{4}} \cos^4 x dx$$
Using the identity $\cos^4 x = \left(\frac{1 + \cos 2x}{2}\right)^2 = \frac{1}{4}(1 + 2\cos 2x + \cos^2 2x) = \frac{1}{4}(1 + 2\cos 2x + \frac{1 + \cos 4x}{2})$:
$$\cos^4 x = \frac{1}{8}(3 + 4\cos 2x + \cos 4x)$$
Substituting this into the integral:
$$I = 32 \cdot \frac{1}{8} \int_{0}^{\frac{\pi}{4}} (3 + 4\cos 2x + \cos 4x) dx = 4 \left[ 3x + 2\sin 2x + \frac{\sin 4x}{4} \right]_0^{\frac{\pi}{4}}$$
Evaluating at the limits:
$$I = 4 \left( [3(\frac{\pi}{4}) + 2\sin(\frac{\pi}{2}) + \frac{\sin \pi}{4}] - [0] \right) = 4 \left( \frac{3\pi}{4} + 2 + 0 \right) = 3\pi + 8$$
The value of the integral is $3\pi + 8$.