Derivatives & Antiderivatives (College Board AP® Calculus BC): Revision Note

Roger B

Written by: Roger B

Reviewed by: Dan Finlay

Updated on

Derivatives & antiderivatives

How can I find the indefinite integrals of common functions?

  • Because differentiation and integration are inverse operations

    • you can 'reverse' what you know about derivatives to find indefinite integrals

  • I.e. if f'(x)=g(x)

    • then g(x) dx=f(x)+C

  • This means that all your derivative results for common functions

    • have indefinite integral equivalents

Indefinite integrals of powers of x

  • ddx(xn)=nxn1 therefore

    • xn dx=1n+1xn+1+C,  n1

      • Note that you can't integrate x1=1x using this rule

        • The denominator in the fraction would become zero

        • 1x must be integrated using logarithms

  • Also note these two special cases

    • k dx=kx+C, where k is a constant

    • 0 dx=C

Worked Example

Find the indefinite integral x10 dx.

Answer:

Use xn dx=1n+1xn+1+C,  n1

x10 dx=110+1x10+1+C

x10 dx=111x11+C

Indefinite integrals of exponentials and 1/x

  • ddx(ex)=ex  and  ddx(ekx)=kekx  therefore

    • ex dx=ex+C

    • ekx dx=1kekx+C

  • ddx(ax)=ax lna  and  ddx(akx)=akx k lna  therefore

    • ax dx=1lna ax+C

    • akx dx=1klna akx+C

  • ddx(lnx)=1x  therefore

    • 1x dx=ln|x|+C

      • Don't forget the modulus (absolute value) sign around the x

      • This allows the integral to be valid for negative values of x as well as for positive values

  • In the above formulas, k is a real number constant and a is a positive real number constant

Worked Example

Find the indefinite integral e7x dx.

Answer:

Use ekx dx=1kekx+C

e7x dx=17e7x+C

Indefinite integrals of trigonometric functions

  • ddx(sinx)=cosx  and  ddx(sinkx)=kcoskx  therefore

    • cosx dx=sinx+C

    • coskx dx=1ksinkx+C

  • ddx(cosx)=sinx  and  ddx(coskx)=ksinkx  therefore

    • sinx dx=cosx+C

    • sinkx dx=1kcoskx+C

  • ddx(tanx)=sec2x  and  ddx(tankx)=ksec2kx  therefore

    • sec2x dx=tanx+C

    • sec2kx dx=1ktankx+C

  • In the above formulas, k is a real number constant

Worked Example

Find the following indefinite integrals:

(a) sin4x dx

(b) (1cos3x)2 dx

Answer:

(a)

Use sinkx dx=1kcoskx+C

sin4x dx=14cos4x+C

(b)

Remember secx=1cosx

(1cos3x)2=(sec3x)2=sec23x

Use sec2kx dx=1ktankx+C

(1cos3x)2 dx=sec23x dx=13tan3x+C

Indefinite integrals of reciprocal trigonometric functions

  • ddx(secx)=tanx secx  and  ddx(seckx)=ktankx seckx  therefore

    • tanx secx dx=secx+C

    • tankx seckx dx=1kseckx+C

  • ddx(cscx)=cotx cscx  and  ddx(csckx)=kcotkx csckx  therefore

    • cotx cscx dx=cscx+C

    • cotkx csckx dx=1kcsckx+C

  • ddx(cotx)=csc2x  and  ddx(cotkx)=kcsc2kx  therefore

    • csc2x dx=cotx+C

    • csc2kx dx=1kcotkx+C

  • In the above formulas, k is a real number constant

Worked Example

Find the indefinite integral cot5x csc5x dx

Answer:

Use cotkx csckx dx=1kcsckx+C

cot5x csc5x dx=15csc5x+C

Indefinite integrals using inverse trigonometric functions

  • ddx(arcsinx)=11x2,  1<x<1,  therefore

    • 11x2 dx=arcsinx+C,  1<x<1

    • 1k2x2 dx=arcsin(xk)+C,  k<x<k

  • ddx(arccosx)=11x2,  1<x<1,  therefore

    • 11x2 dx=arccosx+C,  1<x<1

  • You can see that 11x2 dx is either arcsinx+C or arccosx+C

    • Usually arcsin is used when finding indefinite integrals of this form

  • ddx(arctanx)=11+x2  therefore

    • 11+x2 dx=arctanx+C

    • 1k2+x2 dx=1karctan(xk)+C

Table of common indefinite integrals

In the table below, k is a real number constant and a is a positive real number constant

Standard derivative

Corresponding indefinite integral

ddx(xn)=nxn1

xn dx=1n+1xn+1+C,  n1

ddx(kx)=k

k dx=kx+C

derivative of a constant is zero

0 dx=C

 ddx(ekx)=kekx 

ekx dx=1kekx+C

 ddx(akx)=akx k lna 

akx dx=1klna akx+C

ddx(lnx)=1x 

1x dx=ln|x|+C

 ddx(sinkx)=kcoskx 

coskx dx=1ksinkx+C

 ddx(coskx)=ksinkx 

sinkx dx=1kcoskx+C

 ddx(tankx)=ksec2kx 

sec2kx dx=1ktankx+C

 ddx(seckx)=ktankx seckx 

tankx seckx dx=1kseckx+C

 ddx(csckx)=kcotkx csckx 

cotkx csckx dx=1kcsckx+C

 ddx(cotkx)=kcsc2kx 

csc2kx dx=1kcotkx+C

ddx(arcsinx)=11x2,  1<x<1

11x2 dx=arcsinx+C,  1<x<1

ddx(arctanx)=11+x2 

11+x2 dx=arctanx+C

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Roger B

Author: Roger B

Expertise: Development Editor

Roger's teaching experience stretches all the way back to 1992, and in that time he has taught students at all levels between Year 7 and university undergraduate. Having conducted and published postgraduate research into the mathematical theory behind quantum computing, he is more than confident in dealing with mathematics at any level the exam boards might throw at you.

Dan Finlay

Reviewer: Dan Finlay

Expertise: Portfolio Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.