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

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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