Question

The integral \( \int e^x \frac{2 + \sin 2x}{1 + \cos 2x} \, dx \) is equal to

• 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:

When dealing with integrals involving trigonometric functions, it’s useful to apply identities to simplify the integrand before attempting the integration.

Answer 1

The Correct Option is B

We are given the integral: \[ I = \int e^x \frac{2 + \sin 2x}{1 + \cos 2x} \, dx. \] Step 1: Simplify the expression inside the integral. Recall the trigonometric identity: \[ 1 + \cos 2x = 2\cos^2 x. \] Substitute this into the denominator: \[ I = \int e^x \frac{2 + \sin 2x}{2\cos^2 x} \, dx. \] Next, separate the terms: \[ I = \int e^x \frac{2}{2\cos^2 x} \, dx + \int e^x \frac{\sin 2x}{2\cos^2 x} \, dx. \] Simplify the fractions: \[ I = \int e^x \sec^2 x \, dx + \int e^x \tan x \sec x \, dx. \] Step 2: Evaluate each integral. For the first term: \[ \int e^x \sec^2 x \, dx, \] we use the standard result: \[ \int e^x \sec^2 x \, dx = e^x \tan x + C_1. \] For the second term: \[ \int e^x \tan x \sec x \, dx, \] we apply the known formula: \[ \int e^x \tan x \sec x \, dx = e^x \tan x + C_2. \] Step 3: Combine the results. Adding both terms together: \[ I = e^x \tan x + C. \] Final Answer: \[ \boxed{e^x \tan x + C} \]