Question

Evaluate the integral: $ I = \int e^{2x+3} \sin 6x \, dx $.

• A. \( \frac{e^{2x+3}}{40} (2\sin 6x + 6\cos 6x) + C \)
• B. \( \frac{e^{2x+3}}{40} (2\cos 6x + 6\sin 6x) + C \)
• C. \( \frac{e^{2x+3}}{20} (\sin 6x - 3\cos 6x) + C \)
• D. \( \frac{e^{2x+3}}{20} (\cos 6x - 3\sin 6x) + C \)

Hint:

For integrals involving \( e^{ax} \sin bx \) or \( e^{ax} \cos bx \), use the standard integration formula to directly obtain the result.

Answer 1

The Correct Option is C

Step 1: Using the Standard Integral Formula For integrals of the form: \[ \int e^{ax} \sin(bx) \, dx \] we use the standard formula: \[ \int e^{ax} \sin(bx) \, dx = \frac{e^{ax}}{a^2 + b^2} (a \sin bx - b \cos bx). \] Step 2: Identifying Constants Here, we have \( a = 2 \) and \( b = 6 \), so: \[ I = \int e^{2x+3} \sin 6x \, dx. \] Since \( e^{2x+3} = e^3 \cdot e^{2x} \), we factor out \( e^3 \) and apply the formula: \[ I = e^3 \int e^{2x} \sin 6x \, dx. \] Using the formula: \[ I = \frac{e^{2x+3}}{2^2 + 6^2} (2 \sin 6x - 6 \cos 6x). \] Step 3: Evaluating the Denominator \[ 2^2 + 6^2 = 4 + 36 = 40. \] Thus, \[ I = \frac{e^{2x+3}}{40} (2 \sin 6x - 6 \cos 6x). \] Simplifying, \[ I = \frac{e^{2x+3}}{20} (\sin 6x - 3\cos 6x) + C. \]