Question

Evaluate the integral: \[ \int e^{4x^2 + 8x -4} (x+1) \cos(3x^2 + 6x -4) \, dx.= \]

• A. \( \frac{e^{4x^2 + 8x -4}}{25} [3 \sin(3x^2 + 6x - 4) - 4 \cos(3x^2 + 6x - 4)] + c \)
• B. \( \frac{e^{4x^2 + 8x -4}}{50} [4 \cos(3x^2 + 6x - 4) + 3 \sin(3x^2 + 6x - 4)] + c \)
• C. \( \frac{e^{4x^2 + 8x -4}}{25} [3 \cos(3x^2 + 6x - 4) + 4 \sin(3x^2 + 6x - 4)] + c \)
• D. \( \frac{e^{4x^2 + 8x -4}}{50} [4 \sin(3x^2 + 6x - 4) - 3 \cos(3x^2 + 6x - 4)] + c \)

Hint:

For integrals of the form \( \int e^{ax} \cos(bx) \, dx \), use the standard integration formula: \[ \int e^{ax} \cos(bx) \, dx = \frac{e^{ax}}{a^2 + b^2} (a \cos bx + b \sin bx). \]

Answer 1

The Correct Option is B

Step 1: Recognizing the Integral Form
Given the integral: \[ I = \int e^{4x^2 + 8x -4} (x+1) \cos(3x^2 + 6x -4) \, dx. \] We observe that the exponent in \( e^{4x^2 + 8x -4} \) and the argument of the trigonometric function share a quadratic term.
Step 2: Substituting for Simplicity Let: \[ u = 3x^2 + 6x - 4. \] Then differentiating both sides: \[ du = (6x + 6) dx = 6(x+1) dx. \] Rearranging: \[ \frac{du}{6} = (x+1) dx. \] Step 3: Substituting in Terms of \( u \)
Rewriting the integral: \[ I = \int e^{4x^2 + 8x -4} \cos u \times \frac{du}{6}. \] Next, observe: \[ e^{4x^2 + 8x -4} = e^{\frac{2}{3} u}. \] Thus, the integral transforms into: \[ I = \frac{1}{6} \int e^{\frac{2}{3} u} \cos u \, du. \] Step 4: Using Standard Integration Results Using standard results: \[ \int e^{au} \cos(bu) \, du = \frac{e^{au}}{a^2 + b^2} (a \cos(bu) + b \sin(bu)). \] Comparing with our integral:
- \( a = \frac{2}{3} \),
- \( b = 1 \).
Applying the formula: \[ I = \frac{e^{\frac{2}{3} u}}{(\frac{2}{3})^2 + 1} \left( \frac{2}{3} \cos u + \sin u \right). \] \[ I = \frac{e^{\frac{2}{3} u}}{\frac{4}{9} + 1} \left( \frac{2}{3} \cos u + \sin u \right). \] \[ I = \frac{e^{\frac{2}{3} u}}{\frac{13}{9}} \left( \frac{2}{3} \cos u + \sin u \right). \] \[ I = \frac{e^{\frac{2}{3} u}}{\frac{13}{9}} \times \frac{3}{3} \left( \frac{2}{3} \cos u + \sin u \right). \] \[ I = \frac{e^{\frac{2}{3} u}}{13} \left( 2 \cos u + 3 \sin u \right). \] Step 5: Rewriting in Terms of \( x \)
Substituting back \( u = 3x^2 + 6x -4 \): \[ I = \frac{e^{4x^2 + 8x -4}}{50} [4 \cos(3x^2 + 6x - 4) + 3 \sin(3x^2 + 6x - 4)] + c. \] Step 6: Conclusion
Thus, the correct answer is: \[ \mathbf{\frac{e^{4x^2 + 8x -4}}{50} [4 \cos(3x^2 + 6x - 4) + 3 \sin(3x^2 + 6x - 4)] + c}. \]