Question

Integrate the function: $\frac {1}{cos(x+a)cos(x+b)}$

Answer 1

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

Multiplying and dividing by $sin(a-b)$, we obtain

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

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

= \frac {1} {sin(a-b)}$$[\frac {sin(x+a).cos(x+b)-cos(x+a)sin(x+b)}{cos(x+a)cos(x+b)}]

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

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

∫$$\frac {1}{cos(x+a)cos(x+b)}dx= \frac {1} {sin(a-b)}$$∫[tan(x+a)-tan(x+b)]dx

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

                                         =$\frac {1} {sin(a-b)}$$log|\frac {cos(x+b)}{cos(x+a)}|+C$