Question

Evaluate the integral: \[ \int e^{-2x} \left( \tan 2x - 2\sec^2 2x \tan 2x \right) dx. \]

• A. \( e^{-2x} \tan 2x + c \)
• B. \( -\frac{e^{-2x}}{2} \left[ \sec^2 2x + \tan 2x \right] + c \)
• C. \( -\frac{e^{-2x}}{2} \left[ \tan 2x - \sec^2 2x \right] + c \)
• D. \( -e^{-2x} \sec^2 2x + c \)

Hint:

Use substitution when dealing with trigonometric integrals. - Recognizing derivative patterns helps in solving quickly.

Answer 1

The Correct Option is B

Step 1: Use integration by parts
Let: \[ I = \int e^{-2x} \left( \tan 2x - 2\sec^2 2x \tan 2x \right) dx. \] Using substitution \( u = \tan 2x \), so that \( du = 2\sec^2 2x dx \): \[ I = \int e^{-2x} (u - 2 du). \] Step 2: Integrate
\[ \int e^{-2x} u dx - 2\int e^{-2x} du. \] Solving, \[ I = -\frac{e^{-2x}}{2} \left[ \sec^2 2x + \tan 2x \right] + c. \] Thus, the correct answer is \( \boxed{-\frac{e^{-2x}}{2} \left[ \sec^2 2x + \tan 2x \right] + c} \).