allery$\int \sec x\,dx$ 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: syllabus Mathematics Ch 7 "Integrals", p.301_
Keep practising Probability & Statistics
Browse all Probability & Statistics questions for AP Intermediate 1st Year, or explore more Mathematics topics.
Related questions
$\int x\cos x\,dx$ equalsThe integration-by-parts formula $\int u\,\dfrac{dv}{dx}\,dx$ equals$\int \dfrac{dx}{x(x^2+1)}$ equals$\int \dfrac{x\,dx}{(x-1)(x-2)}$ equalsIn decomposing $\dfrac{1}{(x+1)(x+2)}=\dfrac{A}{x+1}+\dfrac{B}{x+2}$, the constants are$\int \dfrac{dx}{(x+1)(x+2)}$ equals$\int \dfrac{dx}{x^2+2x+2}$ equals$\int \dfrac{dx}{x^2-6x+13}$ equals