Question

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

Answer 1

$\frac{1}{Cos(x-a).Cos(x-b)}=\frac{2}{Cos[(x-a)+(x-b)]+Cos[(x-a)-(x-b)]}\\ \\=\frac{2}{Cos(2x-a-b)+Cos(b-a)},\ \ \ \ \ \ Let\ \ 2x-a-b=2y,\ \ \ 2\ dx=2dy\\ \\=\frac{2}{k+Cos\ 2y}, \ \ \ \ Letting\ \ Cos(b-a)=k, -1 <= k <= 1\\ \\ \int\limits^{}_{} {\frac{1}{Cos(x-a).Cos(x-b)}} \, dx = \int\limits^{}_{} {\frac{2}{k+cos\ 2y}} \, dy = I\\ \\Let\ \ tan\ y = t,\ \ \ \ \ Sec^2y\ dy = dt,\ \ \ dy = \frac{dt}{1+t^2}\\ \\I = \int\limits^{}_{} {\frac{2}{[k+\frac{1-t^2}{1+t^2}] (1+t^2)} } \, dt$


$= \int\limits^{}_{} {\frac{2}{k+kt^2+1-t^2} } \, dt \\ \\=\int\limits^{}_{} {\frac{2}{(k+1)-t^2(1-k)} } \, dt=\int\limits^{}_{} {\frac{2/(k+1)}{1-t^2(1-k)/(k+1)} } \, dt\\ \\Let\ \ \ z=t\sqrt{\frac{1-k}{k+1}},\ \ \ dz=dt\ \sqrt{\frac{1-k}{k+1}}$


$I = \int\limits^{}_{} {\frac{2/(k+1)}{1-z^2} } \, dz \sqrt{\frac{k+1}{1-k}}=P\int\limits^{}_{} {\frac{1}{1-z^2} } \, dz ,\ \ \. P = 2\sqrt{\frac{1}{1-k^2}}=\frac{2}{|Sin(b-a)|}$


$I = P\ tanh^{-1}z\\ \\.\ \ \ \ \ \ where\ z = \sqrt{\frac{1-k}{k+1}}\ tan\ (x-\frac{a+b}{2})\\ \\z=\sqrt{\frac{1-k^2}{(k+1)^2}}\ tanh^{-1}\ (x-\frac{a+b}{2})\\ \\I=\frac{2}{|Sin(b-a)|} . tanh^{-1} [|\frac{Sin(b-a)}{1+Cos(b-a)}| (x-\frac{a+b}{2}) ]$