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  • Integrate the functions in Exercises 1 to 24. 11

Integrate the functions in Exercises 1 to 24.

    Q11.    \frac{1}{\cos(x+a)\cos(x+b)}

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For simplifying the given equation, we need to multiply and divide the expression by   \sin (a-b).

Thus we obtain :

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

or                                                                                           = \frac{1}{ \sin (a-b)}\times \frac{\sin{\left [ (x+a) - (x+b) \right ]}}{\cos (x+a) \cos (x+b)}

or                                                                                            = \frac{1}{ \sin (a-b)}\times \left ( \frac{\sin (x+a) }{\cos (x+a) } - \frac{\sin(x+b)}{\cos (x+b)} \right )

or                                                                                             = \frac{1}{ \sin (a-b)}\times \left ( \tan(x+a)\ -\ \tan(x+b) \right )

Thus integral becomes  :

                                               \int \frac{1}{\cos(x+a)\cos(x+b)}\ =\ \frac{1}{\sin (a-b)} \times \int \left ( \tan(x+a)\ -\ \tan(x+b) \right )dx

or                                                                                                  =\ \frac{1}{\sin (a-b)} \times \left [ -\log \left | \cos (x+a) \right | + \log \left | \cos(x+b) \right | \right ]\ +\ C

or                                                                                                   =\ \frac{1}{\sin (a-b)} \times \log \left [ \frac{\cos(x+b) }{cos(x+a)} \right ]\ +\ C

Posted by

Devendra Khairwa

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