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Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 25 Maths Textbook Solution.

Answers (1)

Answer:

\frac{1}{\sin (a-b)} \log |\sec (x-b)|-\log |\sec (x-a)|+c

Given:

\int \frac{1}{\cos (x-a) \cos (x-b)} d x

Hint:

To solve this statement we have to convert cos into tan.

Solution: 

\frac{1}{\cos (x-a) \cos (x-b)}=\frac{1}{\sin (a-b)} \frac{\sin (x-b)-(x-a)}{\cos (x-a) \cos (x-b)}

=\frac{1}{\sin (a-b)} \frac{\sin (x-b) \cdot \cos (x-a)-\sin (x-a) \cos (x-b)}{\cos (x-a) \cos (x-b)}

=\frac{1}{\sin (a-b)} \frac{\sin (x-b)}{\cos (x-b)}-\frac{\sin (x-a)}{\cos (x-a)}

=\frac{1}{\sin (a-b)} \cdot \tan (x-b)-\tan (x-a)

I=\int \frac{1}{\cos (x-a) \cos (x-b)} d x

   =\frac{1}{\sin (a-b)} \int[\tan (x-b)-\tan (x-a)] d x

   =\frac{1}{\sin (a-b)} \log |\sec (x-b)|-\log |\sec (x-a)|+c

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