Question

Find: \[ \int \frac{2 + \sin 2x}{1 + \cos 2x} e^x \, dx \]

Hint:

For integrals involving trigonometric expressions, simplify using standard identities before integrating.

Answer 1

1. Simplify the trigonometric expression: Using the identity \( 1 + \cos 2x = 2\cos^2 x \) and \( \sin 2x = 2\sin x \cos x \), rewrite the numerator: \[ \frac{2 + \sin 2x}{1 + \cos 2x} = \frac{2 + 2\sin x \cos x}{2\cos^2 x} = \frac{1}{\cos^2 x} + \frac{\sin x}{\cos x} = \sec^2 x + \tan x. \] 2.Rewrite the integral: \[ \int \frac{2 + \sin 2x}{1 + \cos 2x} e^x \, dx = \int (\sec^2 x + \tan x) e^x \, dx. \] 3. Separate the terms: \[ \int (\sec^2 x + \tan x) e^x \, dx = \int \sec^2 x e^x \, dx + \int \tan x e^x \, dx. \] 4. Solve the integrals: The first integral: \[ \int \sec^2 x e^x \, dx = e^x \tan x + C_1. \] The second integral: \[ \int \tan x e^x \, dx = e^x \ln|\sec x| + C_2. \] 5. Combine the results: \[ \int \frac{2 + \sin 2x}{1 + \cos 2x} e^x \, dx = e^x (\tan x + \ln|\sec x|) + C. \] Final Answer: \[ \boxed{e^x (\tan x + \ln|\sec x|) + C.} \]