Question

$\int \frac{1}{\cos \left(x+a\right)\cos \left(x+b\right)}\mathrm{d}x$

Answer 1

​$$\begin{gathered} \int \frac{1}{\cos \left(x+a\right)\cos \left(x+b\right)}\mathrm{d}x \\ \mathrm{Multiplying}\mathrm{and}\mathrm{Dividing}\mathrm{by}\sin \left[\left(x+b\right)-\left(x+a\right)\right],\mathrm{we}\mathrm{get} \\ =\int \frac{1}{\sin \left[\left(x+b\right)-\left(x+a\right)\right]}\times \frac{\sin \left[\left(x+b\right)-\left(x+a\right)\right]}{\cos \left(x+a\right)\cos \left(x+b\right)}\mathrm{d}x \\ =\int \frac{1}{\sin \left(b-a\right)}\times \frac{\sin \left[\left(x+b\right)-\left(x+a\right)\right]}{\cos \left(x+a\right)\cos \left(x+b\right)}\mathrm{d}x \\ =\frac{1}{\sin \left(b-a\right)}\int \frac{\sin \left(x+b\right)\cos \left(x+a\right)-\sin \left(x+a\right)\cos \left(x+b\right)}{\cos \left(x+a\right)\cos \left(x+b\right)}\mathrm{d}x \\ =\frac{1}{\sin \left(b-a\right)}\left[\int \frac{\sin \left(x+b\right)}{\cos \left(x+b\right)}\mathrm{d}x-\int \frac{\sin \left(x+a\right)}{\cos \left(x+a\right)}\mathrm{d}x\right] \\ =\frac{1}{\sin \left(b-a\right)}\left[\int \tan \left(x+b\right)\mathrm{d}x-\int \tan \left(x+a\right)\mathrm{d}x\right] \\ =\frac{1}{\sin \left(b-a\right)}\left[\log \left(\sec \left(x+b\right)\right)-\log \left(\sec \left(x+a\right)\right)\right]+c \\ =\frac{1}{\sin \left(b-a\right)}\left[\log \left(\frac{\sec \left(x+b\right)}{\sec \left(x+a\right)}\right)\right]+c \end{gathered}$$

Hence, $\int \frac{1}{\cos \left(x+a\right)\cos \left(x+b\right)}\mathrm{d}x=\frac{1}{\sin \left(b-a\right)}\left[\log \left(\frac{\sec \left(x+b\right)}{\sec \left(x+a\right)}\right)\right]+c$.