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प्रश्न
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उत्तर
\[\int\frac{1}{\cos 3x - \cos x}dx\]
\[ = \int\frac{1}{4 \cos^3 x - 4\ cosx}dx \left[ \because \cos 3x = 4 \cos {}^3 x - 3 \cos x \right]\]
\[ = \int\frac{1}{4\cos x\left( \cos^2 x - 1 \right)}dx\]
\[ = \frac{- 1}{4}\int\frac{1}{\cos x \sin^2 x}dx\]
\[ = \frac{- 1}{4}\int\left( \frac{\sin^2 x + \cos^2 x}{\cos x \sin^2 x}dx \right)\]
\[ = \frac{- 1}{4}\left[ \int\sec\ x\ dx + \int\text{cot x cosec}\ x\ dx \right]\]
\[ = \frac{- 1}{4}\left( \ln \left| \ sec\ x + \ tan\ x \right| - cosec\ x \right) + C\]
\[ = \frac{1}{4}\left( \text{cosec x} - \ln\left| \sec x + \tan x \right| \right) + C\]
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