Question:
Evaluate the integral:
\[
\int \frac{2\cos 3x - 3\sin 3x}{\cos 3x + 2\sin 3x} dx.
\]
Evaluate the integral:
\[
\int \frac{2\cos 3x - 3\sin 3x}{\cos 3x + 2\sin 3x} dx.
\]
Show Hint
Use substitution to simplify trigonometric expressions.
- Logarithmic integration is common for rational trigonometric fractions.
Updated On: Mar 6, 2026
- \( \frac{7}{15} \log |\cos 3x + 2\sin 3x| - \frac{4}{5} x + c \)
- \( -\frac{4}{5} \log |\cos 3x + 2\sin 3x| + \frac{7x}{5} + c \)
- \( \frac{7}{5} \log |\cos 3x + 2\sin 3x| - \frac{4}{5} x + c \)
- \( -\frac{8}{15} \log |\cos 3x + 2\sin 3x| + \frac{x}{5} + c \)
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The Correct Option is A
Solution and Explanation
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 \( \boxed{\frac{7}{15} \log |\cos 3x + 2\sin 3x| - \frac{4}{5} x + c} \).
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