Question

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

Answer 1

$$\begin{gathered} \int \frac{1}{\cos \left(x-a\right)·\cos \left(x-b\right)}dx \\ =\frac{1}{\sin \left(a-b\right)}\int \frac{\sin \left(a-b\right)}{\cos \left(x-a\right)·\cos \left(x-b\right)}dx \\ =\frac{1}{\sin \left(a-b\right)}\int \frac{\sin \left[\left(x-b\right)-\left(x-a\right)\right]}{\cos \left(x-a\right)·\cos \left(x-b\right)}dx \\ =\frac{1}{\sin \left(a-b\right)}\int \frac{\sin \left(x-b\right)·\cos \left(x-a\right)-\cos \left(x-b\right)·\sin \left(x-a\right)}{\cos \left(x-a\right)·\cos \left(x-b\right)} \\ =\frac{1}{\sin \left(a-b\right)}\int \left[\frac{\sin \left(x-b\right)·\cos \left(x-a\right)}{\cos \left(x-a\right)·\cos \left(x-b\right)}-\frac{\cos \left(x-b\right)·\sin \left(x-a\right)}{\cos \left(x-a\right)·\cos \left(x-b\right)}\right]dx \\ =\frac{1}{\sin \left(a-b\right)}\int \left[\tan \left(x-b\right)-\tan \left(x-a\right)\right]dx \\ =\frac{1}{\sin \left(a-b\right)}\int \tan \left(x-b\right)dx-\int \tan \left(x-a\right)dx \\ =\frac{1}{\sin \left(a-b\right)}\left[\ln \left|\sec \left(x-b\right)\right|-\ln \left|\sec \left(x-a\right)\right|\right]+C \\ \mathit{=}\frac{\mathit{1}}{\sin \mathit{}\left(a\mathit{-}b\right)}\left[\ln \mathit{}\left|\cos \mathit{}\left(x\mathit{-}a\right)\right|\mathit{-}\ln \mathit{}\left|\cos \mathit{}\left(x\mathit{-}b\right)\right|\mathit{}\right]+C \\ =\frac{1}{\sin \left(a-b\right)}\left[\ln \left|\frac{\cos \left(x-a\right)}{\cos \left(x-b\right)}\right|\right]+C \end{gathered}$$