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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evalute the following integrals:
$\int\frac{1}{\cos(\text{x}+\text{a})\cos(\text{x}+\text{b})}\text{dx}$
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Answer

$\int\frac{1}{\cos(\text{x}+\text{a})\cos(\text{x}+\text{b})}\text{dx}$
Multiplying and Dividing by $\sin\big[(\text{x}+\text{b})-(\text{x}+\text{a})\big],$ we get
$=\int\frac{1}{\sin\big[(\text{x}+\text{b})-(\text{x}+\text{a})\big]}\times\frac{\sin\big[(\text{x}+\text{b})-(\text{x}+\text{a})\big]}{\cos(\text{x}+\text{a})\cos(\text{x}+\text{b})}\text{dx}$
$=\int\frac{1}{\sin(\text{b}-\text{a})}\times\frac{\sin\big[(\text{x}+\text{b})-(\text{x}+\text{a})\big]}{\cos(\text{x}+\text{a})\cos(\text{x}+\text{b})}\text{dx}$
$=\frac{1}{\sin(\text{b}-\text{a})}\int\frac{\sin(\text{x}+\text{b})\cos(\text{x}+\text{a})-\sin(\text{x}+\text{a})\cos(\text{x}+\text{b})}{\cos(\text{x}+\text{a})\cos(\text{x}+\text{b})}\text{dx}$
$=\frac{1}{\sin(\text{b}-\text{a})}\Big[\int\frac{\sin(\text{x}+\text{b})}{\cos(\text{x}+\text{b})}\text{dx}-\int\frac{\sin(\text{x}+\text{a})}{\cos(\text{x}+\text{a})}\text{dx}\Big]$
$=\frac{1}{\sin(\text{b}-\text{a})}\Big[\int\tan(\text{x}+\text{b})\text{dx}-\int\tan(\text{x}+\text{a})\text{dx}\Big]$
$=\frac{1}{\sin(\text{b}-\text{a})}\big[\log(\sec(\text{x}+\text{b}))-\log(\sec(\text{x}+\text{a}))\big]+\text{C}$
$=\frac{1}{\sin(\text{b}-\text{a})}\bigg[\log\Big(\frac{\sec(\text{x}+\text{b})}{\sec(\text{x}+\text{a})}\Big)\bigg]+\text{C}$
Hence, $\int\frac{1}{\cos(\text{x}+\text{a})\cos(\text{x}+\text{b})}\text{dx}=\frac{1}{\sin(\text{b}-\text{a})}\bigg[\log\Big(\frac{\sec(\text{x}+\text{b})}{\sec(\text{x}+\text{a})}\Big)\bigg]+\text{C}$

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