Question

The value of \[ \int \frac{\sin 7x}{\cos 9x \, \cos 2x} \, dx \] is:

• A. \((\log \sec(9x) - \log \sec(2x) + c) \)
• B. \((\log \sec(9x) + \log \sec(2x) + c) \)
• C. \((\frac{1}{9} \log \sec(9x) - \frac{1}{2} \log \sec(2x) + c) \)
• D. \((\frac{1}{9} \log \sec(9x) + \frac{1}{2} \log \sec(2x) + c)\)

Hint:

Always look to express the numerator's angle as a sum or difference of the denominator's angles.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The integral features trigonometric functions in the numerator and denominator. We aim to express the numerator in terms of the angles in the denominator.
Step 3: Detailed Explanation:
Notice that $7x = 9x - 2x$.
So, $\sin 7x = \sin(9x - 2x) = \sin 9x \cos 2x - \cos 9x \sin 2x$.
Substitute this into the integral:
$I = \int \frac{\sin 9x \cos 2x - \cos 9x \sin 2x}{\cos 9x \cos 2x} dx$
Splitting the fraction:
$I = \int \left( \frac{\sin 9x \cos 2x}{\cos 9x \cos 2x} - \frac{\cos 9x \sin 2x}{\cos 9x \cos 2x} \right) dx$
$I = \int (\tan 9x - \tan 2x) dx$
$I = \int \tan 9x dx - \int \tan 2x dx$
Using the standard integral $\int \tan(ax) dx = \frac{1}{a} \log|\sec ax| + c$:
$I = \frac{1}{9} \log|\sec 9x| - \frac{1}{2} \log|\sec 2x| + c$
Step 4: Final Answer:
The integral is $\frac{1}{9} \log \sec(9x) - \frac{1}{2} \log \sec(2x) + c$.