$\int \sec x\,dx$ 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_
Related questions
When variables and numbers appear on both sides, you must simplify:Equations needing more than one operation are described as taking more:Once a solution is found, it should be checked by substituting it back to get a statement The product of any number and its reciprocal equals:A fraction multiplying the variable is best removed by multiplying by its:Solving x divided by 4 equals 3 gives x equal to:Solving 4x equals 20 gives x equal to:Solving x minus 8 equals 5 gives x equal to: