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MHT CET · Maths · Indefinite Integration

\(\int \frac{\sin 7 x}{\cos 9 x \cos 2 x} \mathrm{~d} x\) is equal to

  1. A \(\log \sec (9 x)-\log \sec (2 x)+c\), where \(c\) is the constant of integration
  2. B \(\log \sec (9 x)+\log \sec (2 x)+c, \quad\) where \(c\) is the constant of integration
  3. C \(\frac{1}{9} \log \sec (9 x)-\frac{1}{2} \log \sec (2 x)+c\), where \(c\) is the constant of integration
  4. D \(\frac{1}{9} \log \sec (9 x)+\frac{1}{2} \log \sec (2 x)+c\), where \(c\) is the constant of integration
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{1}{9} \log \sec (9 x)-\frac{1}{2} \log \sec (2 x)+c\), where \(c\) is the constant of integration

Step-by-step Solution

Detailed explanation

\( \int \frac{\sin (9x - 2x)}{\cos 9x \cos 2x} \mathrm{~d} x \) \( \int \frac{\sin 9x \cos 2x - \cos 9x \sin 2x}{\cos 9x \cos 2x} \mathrm{~d} x \)