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Elementary primitives and exact definite integrals

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Anti-differentiate exponential and circular functions

For an exponential or circular function with a linear argument, divide by the coefficient of the variable. Sine changes to negative cosine, cosine changes to sine, and the exponent of an exponential stays unchanged. These formulas require a nonzero coefficient of the variable. Each indefinite integral includes one arbitrary constant

\[\int e^{kx}\,dx=\frac{1}{k}e^{kx}+C,\quad\int\sin(kx)\,dx=-\frac{1}{k}\cos(kx)+C,\quad\int\cos(kx)\,dx=\frac{1}{k}\sin(kx)+C\]
  1. 1Keep the exponential argument Divide the coefficient by k and leave the exponent kx unchanged \(\displaystyle \int4e^{2x}\,dx=2e^{2x}+C\)
  2. 2Keep the sign of a negative rate A negative coefficient of x produces a negative reciprocal factor \(\displaystyle \int6e^{-3x}\,dx=-2e^{-3x}+C\)
  3. 3Change sine to negative cosine The sine anti-derivative includes both the reciprocal factor and a negative sign \(\displaystyle \int4\sin(2x)\,dx=-2\cos(2x)+C\)
  4. 4Change cosine to sine The cosine anti-derivative changes the function and includes the reciprocal factor \(\displaystyle \int3\cos(3x)\,dx=\sin(3x)+C\)
Year 12, VCE Units 3 & 4, Chapter 8, 4 difficulty levels, No calculatorBuild a worksheet

Anti-differentiate exponential and circular functions

For an exponential or circular function with a linear argument, divide by the coefficient of the variable. Sine changes to negative cosine, cosine changes to sine, and the exponent of an exponential stays unchanged. Here k≠0, and each complete family includes one arbitrary constant

\[\int e^{kx}\,dx=\frac{1}{k}e^{kx}+C,\quad\int\sin(kx)\,dx=-\frac{1}{k}\cos(kx)+C,\quad\int\cos(kx)\,dx=\frac{1}{k}\sin(kx)+C\]
  1. Keep the exponential argument Divide the coefficient by k and leave the exponent kx unchanged \(\displaystyle \int4e^{2x}\,dx=2e^{2x}+C\)
  2. Keep the sign of a negative rate A negative coefficient of x produces a negative reciprocal factor \(\displaystyle \int6e^{-3x}\,dx=-2e^{-3x}+C\)
  3. Change sine to negative cosine The sine anti-derivative includes both the reciprocal factor and a negative sign \(\displaystyle \int4\sin(2x)\,dx=-2\cos(2x)+C\)
  4. Change cosine to sine The cosine anti-derivative changes the function and includes the reciprocal factor \(\displaystyle \int3\cos(3x)\,dx=\sin(3x)+C\)
  5. Use both limits Evaluate the same primitive at b and a, then subtract in the order upper minus lower. Constants cancel \(\displaystyle \int_a^b f(x)\,dx=F(b)-F(a)\)

Worked example

Find the complete family of anti-derivatives of the function below

\[\int6\sin(2x)\,dx\]
  1. \(\displaystyle a=6,\quad k=2\)The coefficient is 6 and the sine argument has inner coefficient 2.
  2. \(\displaystyle \frac{a}{k}=\frac{6}{2}=3\)Divide the coefficient by the inner coefficient
  3. \(\displaystyle \int6\sin(2x)\,dx=-3\cos(2x)+C\)Sine changes to negative cosine, so the coefficient is −3.
  4. \(\displaystyle \frac{d}{dx}\left(-3\cos(2x)+C\right)=6\sin(2x)\)Differentiation returns the original integrand, which checks the result

Common mistakes

Omitting the reciprocal factor
✕\(\displaystyle \int6\sin(2x)\,dx=-6\cos(2x)+C\)✓\(\displaystyle \int6\sin(2x)\,dx=-3\cos(2x)+C\)
The coefficient must be divided by the inner coefficient 2 before the sine sign is applied
Using the wrong circular sign
✕\(\displaystyle \int3\sin(2x)\,dx=\frac{3}{2}\cos(2x)+C\)✓\(\displaystyle \int3\sin(2x)\,dx=-\frac{3}{2}\cos(2x)+C\)
The derivative of cosine is negative sine, so the anti-derivative of sine needs a negative sign
Changing the exponential argument
✕\(\displaystyle \int4e^{2x}\,dx=2e^{4x}+C\)✓\(\displaystyle \int4e^{2x}\,dx=2e^{2x}+C\)
Anti-differentiation changes the coefficient but keeps the exponential argument 2x

Difficulty levels

introRetain single-term primitive practice and add exact one-term two-limit evaluations
standardRetain elementary primitive tasks and add exact exponential combinations, circular surds and π coefficients
challengeRetain mixed primitive tasks and verify a supplied product or composite primitive before evaluating both limits
examMix retained primitive work with exact two-limit coefficient tasks. These are component checks, not a complete written examination response

Questions about this topic

Why is the coefficient divided by k?+

Differentiating e^(kx), sine(kx) or cosine(kx) creates a factor k. Dividing by k removes that extra factor

Why does the anti-derivative of sine have a negative sign?+

The derivative of cos(kx) is −k sin(kx), so −cos(kx) divided by k differentiates to sin(kx)

What happens to the exponent of e^(kx)?+

The exponent remains kx. Only the coefficient is changed by division by k.

Why does a zero lower limit not remove the second substitution?+

A primitive need not equal zero at zero. For example, 3e^(2x) has value 3 at zero, which must still be subtracted

Can I use a calculator?+

No. This topic is practised without a calculator

Which year level and chapter is this?+

Year 12, VCE Units 3 & 4, Mathematical Methods, chapter 8: Integration

In this chapter

Approximate area by an endpoint or trapezium ruleEvaluate and transform a definite integralElementary primitives and exact definite integralsFind complete families of polynomial anti-derivativesThe area function and the fundamental theorem of calculusSigned integrals, geometric area and motionFind the area between two curvesAverage value of a functionTotal change and kinematics from a rateDetermine a polynomial from its derivative and a boundary conditionRecover motion from acceleration