Question

\( \int e^{4x}(\sin 3x - \cos 3x) dx = \)

• A. \( \frac{e^{4x}}{25}(7\sin 3x - \cos 3x) + c \)
• B. \( \frac{e^{4x}}{25}(\sin 3x - 7\cos 3x) + c \)
• C. \( \frac{e^{4x}}{5}(7\sin 3x + \cos 3x) + c \)
• D. \( \frac{e^{4x}}{5}(\sin 3x + 7\cos 3x) + c \)

Hint:

You can also use the method of undetermined coefficients: assume \( I = e^{4x}(A\sin 3x + B\cos 3x) \), differentiate, and solve for A and B.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:

We can use the standard integration formulas for \( \int e^{ax} \sin bx \, dx \) and \( \int e^{ax} \cos bx \, dx \).
Step 2: Key Formula or Approach:

1. \( \int e^{ax} \sin bx \, dx = \frac{e^{ax}}{a^2+b^2}(a \sin bx - b \cos bx) \) 2. \( \int e^{ax} \cos bx \, dx = \frac{e^{ax}}{a^2+b^2}(a \cos bx + b \sin bx) \)
Step 3: Detailed Explanation:

Here \( a = 4 \) and \( b = 3 \). The denominator is \( 4^2 + 3^2 = 25 \). Integral \( I = \int e^{4x}\sin 3x \, dx - \int e^{4x}\cos 3x \, dx \). Applying formulas: \[ I = \left[ \frac{e^{4x}}{25}(4\sin 3x - 3\cos 3x) \right] - \left[ \frac{e^{4x}}{25}(4\cos 3x + 3\sin 3x) \right] \] Factor out \( \frac{e^{4x}}{25} \): \[ I = \frac{e^{4x}}{25} [ (4\sin 3x - 3\cos 3x) - (4\cos 3x + 3\sin 3x) ] \] Group sine and cosine terms: \[ I = \frac{e^{4x}}{25} [ (4-3)\sin 3x + (-3-4)\cos 3x ] \] \[ I = \frac{e^{4x}}{25} [ \sin 3x - 7\cos 3x ] + c \]
Step 4: Final Answer:

Matches Option (B).