allery$\int \tan x\,dx$ equals
A$\log|\cos x|+C$
B$\sec^2 x+C$
C$\log|\sin x|+C$
D$\log|\sec x|+C$
Answer & Solution
Correct answer: D. $\log|\sec x|+C$
1. Write $\tan x=\dfrac{\sin x}{\cos x}$.
2. Put $t=\cos x$, $dt=-\sin x\,dx$.
3. Integral $=-\int \dfrac{dt}{t}=-\log|t|=-\log|\cos x|$.
4. Since $-\log|\cos x|=\log|\sec x|$, the answer is $\log|\sec x|+C$ (option A is the unsimplified negative).
_Source: syllabus Mathematics Ch 7 "Integrals", p.300_
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