Question
Integrate the function in Exercise:
$\frac{1}{\cos(\text{x}+\text{a)}\cos(\text{x}+\text{b)}}$
$\frac{1}{\cos(\text{x}+\text{a)}\cos(\text{x}+\text{b)}}$
Multiplying and dividing by $\sin(\text{a}-\text{b)},$ we obtain
$\frac{1}{\sin\text{(a}-\text{b})}\bigg[\frac{\sin\text{(a}-\text{b})}{\cos\text{(x}+\text{a)}\cos\text{(x}+\text{b)}}\bigg]$
$=\frac{1}{\sin\text{(a}-\text{b})}\bigg[\frac{\sin[\text{(x}+\text{a})-\text{(x}+\text{b)]}}{\cos\text{(x}+\text{a)}\cos\text{(x}+\text{b)}}\bigg]$
$=\frac{1}{\sin\text{(a}-\text{b})}\bigg[\frac{\sin\text{(x}+\text{a}).\cos\text{(x}+\text{b)}-\cos\text{(x}+\text{a)}\sin\text{(x}+\text{b)}}{\cos\text{(x}+\text{a)}\cos\text{(x}+\text{b)}}\bigg]$
$=\frac{1}{\sin\text{(a}-\text{b})}\bigg[\frac{\sin\text{(x}+\text{a)}}{\cos\text{(x}+\text{a)}}-\frac{\sin\text{(x}+\text{b)}}{\cos\text{(x}+\text{b)}}\bigg]$
$=\frac{1}{\sin\text{(a}-\text{b)}}\big[\tan(\text{x}+\text{a)}-\tan\text{(x}+\text{b)}\big]$
$\int\frac{1}{\cos\text{(x}+\text{a)}\cos\text{(x}+\text{b)}}\text{dx}=\frac{1}{\sin\text{(a}-\text{b)}}\int\big[\tan\text{(x}+\text{a)}-\tan\text{(x}+\text{b)}\big]\text{dx}$
$=\frac{1}{\sin\text{(a}-\text{b)}}[-\log|\cos\text{(x}+\text{a)}|+\log|\cos\text{(x}+\text{b)|]}+\text{C}$
$=\frac{1}{\sin\text{(a}-\text{b)}}\log\bigg|\frac{\cos\text{(x}+\text{b)}}{\cos\text{(x}+\text{a)}}\bigg|+\text{C}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.