Question

Evaluate the integral: \[ \int \frac{2\cos 3x - 3\sin 3x}{\cos 3x + 2\sin 3x} dx. \]

• A. \( \frac{7}{15} \log |\cos 3x + 2\sin 3x| - \frac{4}{5} x + c \)
• B. \( -\frac{4}{5} \log |\cos 3x + 2\sin 3x| + \frac{7x}{5} + c \)
• C. \( \frac{7}{5} \log |\cos 3x + 2\sin 3x| - \frac{4}{5} x + c \)
• D. \( -\frac{8}{15} \log |\cos 3x + 2\sin 3x| + \frac{x}{5} + c \)

Hint:

Use substitution to simplify trigonometric expressions. - Logarithmic integration is common for rational trigonometric fractions.

Answer 1

The Correct Option is A

Step 1: Use substitution
Let \( u = \cos 3x + 2\sin 3x \), then differentiate: \[ du = (-3\sin 3x + 6\cos 3x)dx. \] Rewriting, \[ I = \int \frac{du}{u} - \frac{4}{5} \int dx. \] Step 2: Compute the integral
\[ I = \frac{7}{15} \log |u| - \frac{4}{5} x + c. \] Thus, the correct answer is \( \boxed{\frac{7}{15} \log |\cos 3x + 2\sin 3x| - \frac{4}{5} x + c} \).