$\int \sec x\,dx$ equals
A$\log|\sec x|+C$
B$\log|\operatorname{cosec} x-\cot x|+C$
C$\log|\sec x+\tan x|+C$
D$\sec x\tan x+C$
Answer & Solution
Correct answer: C. $\log|\sec x+\tan x|+C$
1. Multiply and divide by $(\sec x+\tan x)$.
2. Put $t=\sec x+\tan x$, so $dt=\sec x(\tan x+\sec x)\,dx$.
3. Integral $=\int \dfrac{dt}{t}=\log|t|$.
4. Hence $\int \sec x\,dx=\log|\sec x+\tan x|+C$. Option B is the result for $\operatorname{cosec} x$.
_Source: NCERT Class 12 Mathematics Ch 7 "Integrals", p.301_
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