Integrate the function 1 (x+a) (x+b)

Question

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

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Answer

Given function is: $\frac{1}{\cos (x+a) \cos (x+b)}$
Let $I=\frac{1}{\cos (x+a) \cos (x+b)}$
Multiply and divide by sin (a - b), we get
$I=\frac{1}{\sin (a-b)} \cdot\left(\frac{\sin (a-b)}{\cos (x+a) \cos (x+b)}\right)$
= $\frac{1}{\sin (a-b)} \cdot\left(\frac{\sin (a-b+x-x)}{\cos (x+a) \cos (x+b)}\right)$
= $\frac{1}{\sin (a-b)} \cdot\left(\frac{\sin [(x+a)-(x+b)]}{\cos (x+a) \cos (x+b)}\right)$
As, {sin (A - B) = sin A cos B - cos A sin B}
$\Rightarrow$ I = $\frac{1}{\sin (a-b)} \cdot\left(\frac{\sin (x+a) \cdot \cos (x+b)-\cos (x+a) \cdot \sin (x+b)}{\cos (x+a) \cos (x+b)}\right)$
= $\frac{1}{\sin (a-b)} \cdot\left(\frac{\sin (x+a) \cdot \cos (x+b)}{\cos (x+a) \cos (x+b)}-\frac{\cos (x+a) \cdot \sin (x+b)}{\cos (x+a) \cos (x+b)}\right)$
= $\frac{1}{\sin (a-b)} \cdot\left(\frac{\sin (x+a)}{\cos (x+a)}-\frac{\sin (x+b)}{\cos (x+b)}\right)$
= $\frac{1}{\sin (a-b)} \cdot[\tan (x+a)-\tan (x+b)]$
$\Rightarrow \int \frac{1}{\cos (x+a) \cos (x+b)} d x$ = $\int \frac{1}{\sin (a-b)} \cdot[\tan (x+a)-\tan (x+b)] d x$
= $\frac{1}{\sin (a-b)}\left\{\int \tan (x+a) d x-\int \tan (x+b) d x\right\}$
= $\frac{1}{\sin (a-b)}[-\log |\cos (x+a)|-(-\log |\cos (x+b)|)]$
= $\frac{1}{\sin (a-b)}[-\log |\cos (x+a)|+\log |\cos (x+b)|]$
$\Rightarrow \mathrm{I}=\frac{1}{\sin (\mathrm{a}-\mathrm{b})} \cdot \log \left|\frac{\cos (\mathrm{x}+\mathrm{b})}{\cos (\mathrm{x}+\mathrm{a})}\right|+\mathrm{c}$

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