Search topics

Search

Anti-differentiate exponential and circular functions

Start a set of 8
Step 3 of 5

Divide the coefficient by the coefficient of x in the sine argument

\[6\sin(2x),\quad\frac{6}{2}=\square\]
Year 12, VCE Units 3 & 4, Chapter 7, 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 the inner coefficient is non-zero, 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 the inner coefficient and leave the exponent 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 \(\displaystyle \int3\cos(3x)\,dx=\sin(3x)+C\)

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

introAnti-differentiate one exponential or circular term with a whole reciprocal factor
standardHandle a signed linear combination, a rule choice or a specific-member condition
challengeAnti-differentiate a mixed exponential and circular integrand or evaluate a conditioned member exactly
examChoose the correct primitive from distractors involving reciprocal factors, signs and function changes

Questions about this topic

Why is the coefficient divided by the inner coefficient?+

Differentiating an exponential, sine or cosine function with a linear argument creates a factor equal to the inner coefficient. Dividing by that coefficient removes the extra factor

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

Differentiating cosine of k times x gives negative k times sine of k times x, so the negative cosine divided by k differentiates to sine of k times x

What happens to the exponent of an exponential with a linear argument?+

The exponent remains unchanged. Only the coefficient is divided by the inner coefficient

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 7: Antiderivatives

In this chapter

7A Antiderivatives and arbitrary constants: Find complete families of polynomial anti-derivatives7B Power rule and integrand preparation: Find complete families of polynomial anti-derivatives7B Power rule and integrand preparation: Recognise equivalent polynomial formsSimplify and combine algebraic fractionsAnti-differentiate a power of a linear function7C Integrating powers of linear functions: Recognise equivalent polynomial forms7C Integrating powers of linear functions: Solve a linear equationAnti-differentiate exponential and circular functionsAnti-differentiate by recognition from a supplied derivativeDetermine a polynomial from its derivative and a boundary conditionInfer primitive curves from a gradient functionReconstruct quantities from their ratesEvaluate and determine function rules7E Antiderivative families and applications: Solve a linear equationRound to decimal places and significant figuresScale a proportional quantity with the unitary method