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: $$\boxed{e^x (\tan x + \ln|\sec x|) + C.}$$