Question

Evaluate the integral: \[ \int \frac{e^x \left( 2 + \sin(2x) \right)}{1 + \cos(2x)} \, dx \]

• A. \( e^x \sec x + C \)
• B. \( e^x \tan x + C \)
• C. \( e^x \cot x + C \)
• D. \( e^x \csc x + C \)

Hint:

Remember to apply trigonometric identities to simplify the expression before integrating.

Answer 1

The Correct Option is B

To evaluate the integral \(\int \frac{e^x \left( 2 + \sin(2x) \right)}{1 + \cos(2x)} \, dx\), we start by simplifying the expression. Use the identity \(1 + \cos(2x) = 2\cos^2(x)\). Thus, the integral becomes:

$$\int \frac{e^x \left( 2 + \sin(2x) \right)}{2\cos^2(x)} \, dx$$

Split the integral:

$$\int \frac{2e^x}{2\cos^2(x)} \, dx + \int \frac{e^x \sin(2x)}{2\cos^2(x)} \, dx$$

Simplifying the first term gives:

$$\int \frac{e^x}{\cos^2(x)} \, dx = \int e^x \sec^2(x) \, dx$$

For the second term, use the identity \(\sin(2x) = 2\sin(x)\cos(x)\), yielding:

$$\int \frac{2e^x \sin(x)\cos(x)}{2\cos^2(x)} \, dx = \int \frac{e^x \sin(x)}{\cos(x)} \, dx = \int e^x \tan(x) \, dx$$

The integral now simplifies to:

$$\int e^x \sec^2(x) \, dx + \int e^x \tan(x) \, dx$$

We solve these using substitution. Let \(u = e^x \tan(x)\). Then, using derivatives \(du = e^x \sec^2(x) \, dx + e^x \tan(x) \, dx\), the integral thus becomes:

$$\int du = u + C = e^x \tan(x) + C$$

Therefore, the evaluated integral is \(e^x \tan x + C\).

Answer 2

We are given the integral: \[ \int \frac{e^x \left( 2 + \sin(2x) \right)}{1 + \cos(2x)} \, dx \] First, simplify the trigonometric expression using trigonometric identities: \[ 1 + \cos(2x) = 2\cos^2(x) \] Thus, the integral becomes: \[ \int \frac{e^x (2 + \sin(2x))}{2 \cos^2(x)} \, dx \] Now, observe that the integrand simplifies further, and using standard integration techniques, the answer is: \[ e^x \tan x + C \]