Question:
\((\int \frac{\sin 7x}{\cos 9x \cos 2x} dx)\) is equal to
\((\int \frac{\sin 7x}{\cos 9x \cos 2x} dx)\) is equal to
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Always look to express the numerator's angle as a sum or difference of the denominator's angles.
Updated On: Apr 30, 2026
\((\log \sec(9x) - \log \sec(2x) + c) \)
\((\log \sec(9x) + \log \sec(2x) + c) \)
\((\frac{1}{9} \log \sec(9x) - \frac{1}{2} \log \sec(2x) + c) \)
\((\frac{1}{9} \log \sec(9x) + \frac{1}{2} \log \sec(2x) + c)\)
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The Correct Option is C
Solution and Explanation
Step 1: Use trigonometric identity
\(Rewrite (\sin 7x = \sin(9x - 2x)). \\ (\sin(9x - 2x) = \sin 9x \cos 2x - \cos 9x \sin 2x). \)
Step 2: Split the fraction
\((\int \frac{\sin 9x \cos 2x}{\cos 9x \cos 2x} , dx - \int \frac{\cos 9x \sin 2x}{\cos 9x \cos 2x} , dx). \\ (\int \tan 9x , dx - \int \tan 2x , dx). \)
Step 3: Integrate
\((\frac{1}{9} \ln|\sec 9x| - \frac{1}{2} \ln|\sec 2x| + c). \)
Final Answer: (C)
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