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INTEGRALS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSINTEGRALS5 Marks
Question
Integrate the function in Exercise:
$\frac{1}{\cos(\text{x}+\text{a)}\cos(\text{x}+\text{b)}}$
✓

Answer

$\frac{1}{\cos(\text{x}+\text{a)}\cos(\text{x}+\text{b)}}$Multiplying and dividing by  $\sin(\text{a}-\text{b)},$ we obtain
$\frac{1}{\sin\text{(a}-\text{b})}\bigg[\frac{\sin\text{(a}-\text{b})}{\cos\text{(x}+\text{a)}\cos\text{(x}+\text{b)}}\bigg]$
$=\frac{1}{\sin\text{(a}-\text{b})}\bigg[\frac{\sin[\text{(x}+\text{a})-\text{(x}+\text{b)]}}{\cos\text{(x}+\text{a)}\cos\text{(x}+\text{b)}}\bigg]$
$=\frac{1}{\sin\text{(a}-\text{b})}\bigg[\frac{\sin\text{(x}+\text{a}).\cos\text{(x}+\text{b)}-\cos\text{(x}+\text{a)}\sin\text{(x}+\text{b)}}{\cos\text{(x}+\text{a)}\cos\text{(x}+\text{b)}}\bigg]$
$=\frac{1}{\sin\text{(a}-\text{b})}\bigg[\frac{\sin\text{(x}+\text{a)}}{\cos\text{(x}+\text{a)}}-\frac{\sin\text{(x}+\text{b)}}{\cos\text{(x}+\text{b)}}\bigg]$
$=\frac{1}{\sin\text{(a}-\text{b)}}\big[\tan(\text{x}+\text{a)}-\tan\text{(x}+\text{b)}\big]$
$\int\frac{1}{\cos\text{(x}+\text{a)}\cos\text{(x}+\text{b)}}\text{dx}=\frac{1}{\sin\text{(a}-\text{b)}}\int\big[\tan\text{(x}+\text{a)}-\tan\text{(x}+\text{b)}\big]\text{dx}$
$=\frac{1}{\sin\text{(a}-\text{b)}}[-\log|\cos\text{(x}+\text{a)}|+\log|\cos\text{(x}+\text{b)|]}+\text{C}$
$=\frac{1}{\sin\text{(a}-\text{b)}}\log\bigg|\frac{\cos\text{(x}+\text{b)}}{\cos\text{(x}+\text{a)}}\bigg|+\text{C}$

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