Question

The integral \[ \int e^x \left( 1 + \tan x + \tan^2 x \right) dx \] is equal to

• A. \( e^x \cos x + c \)
• B. \( e^x \sin x + c \)
• C. \( e^x \tan x + c \)
• D. \( e^x \sec x + c \)

Hint:

When solving integrals involving trigonometric functions, consider breaking the integral into parts and using standard trigonometric identities and integration by parts.

Answer 1

The Correct Option is C

We are given the integral: \[ I = \int e^x \left( 1 + \tan x + \tan^2 x \right) dx \] We can split the integral into separate parts: \[ I = \int e^x \, dx + \int e^x \tan x \, dx + \int e^x \tan^2 x \, dx \] Step 1: Solve the first part \( \int e^x \, dx \) The integral of \( e^x \) is simply: \[ \int e^x \, dx = e^x \] Step 2: Solve the second part \( \int e^x \tan x \, dx \) To solve \( \int e^x \tan x \, dx \), we use integration by parts. Let: - \( u = \tan x \), so \( du = \sec^2 x \, dx \) - \( dv = e^x \, dx \), so \( v = e^x \) Now apply integration by parts: \[ \int e^x \tan x \, dx = e^x \tan x - \int e^x \sec^2 x \, dx \] The second part is just another standard integral, so this simplifies to: \[ \int e^x \tan x \, dx = e^x \tan x - e^x \] Step 3: Solve the third part \( \int e^x \tan^2 x \, dx \) This part can be rewritten as: \[ \int e^x \tan^2 x \, dx = \int e^x (\sec^2 x - 1) \, dx = \int e^x \sec^2 x \, dx - \int e^x \, dx \] We already know the solution to \( \int e^x \, dx \), so we can use the result from earlier: \[ \int e^x \tan^2 x \, dx = e^x \sec^2 x - e^x = e^x (\sec^2 x - 1) \] Step 4: Combine the results Now, combining all parts of the integral: \[ I = e^x + e^x \tan x - e^x + e^x \sec^2 x \] Simplifying: \[ I = e^x \tan x + c \] Thus, the integral is: \[ I = e^x \tan x + c \] Therefore, the correct answer is \( \boxed{e^x \tan x + c} \).