Question

1/cos(x-a) cos(x-b),Integrate it with respect to x.

Answer 1

Answer:

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Answer 2

Answer:

$\mathbf{\frac{1}{sin(b-a)}log\frac{sec(x-a)}{sec(x-b)}+C}$

Step-by-step explanation:

In this question,

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

$\frac{1}{sin(b-a)}\int{\frac{sin{(x-a)-(x-b)}}{cos(x-a)cos(x-b)}}dx$

$\frac{1}{sin(b-a)}\int{\frac{sin(x-a)cos(x-b)-sin(x-b)cos(x-a)}{cos(x-a)cos(x-b)}}dx$

$\frac{1}{sin(b-a)}\int{\frac{sin(x-a)cos(x-b)}{cos(x-a)cos(x-b)}}dx\ -\ \int{\frac{sin(x-b)cos(x-a)}{{cos(x-a)cos(x-b)}}}dx$

$\frac{1}{sin(b-a)}\int{tan(x-a)-tan(x-b)}dx$

$\frac{1}{sin(b-a)}[logsec(x-a)-logsec(x-b)]+C$        

Where C is the arbitrary constant

$\mathbf{\frac{1}{sin(b-a)}log\frac{sec(x-a)}{sec(x-b)}+C}$