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YEAR 12 · INTEGRATION · 16 MIN

Power/exponential/trigonometric antiderivatives

Integrate powers, natural exponentials and basic trigonometric functions, including linear changes of input.

In this lesson

Before you start
  • Power derivatives
  • Exponential and trigonometric derivatives
01 / LEARN

Understand the idea.

Match the derivative pattern

Choose an antiderivative whose derivative matches the whole integrand. For a linear input a⁢x+b, differentiation introduces the constant multiplier a. Integration must divide by a, provided a≠0.

For trigonometric functions, use radians and check signs: cosine differentiates to negative sine, so the antiderivative of sine is negative cosine. The antiderivative of cosine is sine. Exponential functions keep the same input.

∫ea⁢x+bd⁢x=1aea⁢x+b+C(a≠0)

Combine rules carefully

Rewrite roots and reciprocals as powers before choosing a rule: x=x1⁢/2 and 1x2=x−2. Apply each rule only on an interval where the function is defined; the exponent −1 requires the logarithmic rule.

A sum can contain several function types. Integrate its terms independently and add one C. Differentiate the entire answer to check every coefficient, especially when the linear input has a negative slope.

∫cos(a⁢x+b)d⁢x=sin(a⁢x+b)a+C∫sin(a⁢x+b)d⁢x=−cos(a⁢x+b)a+C
02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example

Find ∫(5x4−3ex+2cosx)d⁢x.

Worked solution

  1. 1

    Integrate the power term.

    ∫5x4d⁢x=x5+C1

    Dividing coefficient 5 by the new exponent 5 reverses the power rule.

  2. 2

    Integrate the exponential term.

    ∫(−3ex)d⁢x=−3ex+C2

    The derivative of the natural exponential is itself.

  3. 3

    Integrate the cosine term and combine.

    ∫(5x4−3ex+2cosx)d⁢x=x5−3ex+2sinx+C

    Sine differentiates to cosine, and the original coefficient 2 is retained.

  4. 4

    Check by differentiation.

    dd⁢x⁢(x5−3ex+2sinx+C)=5x4−3ex+2cosx

    All three original terms and signs are recovered.

Answer

x5−3ex+2sinx+C

EXAMPLE 02

Worked example

Find ∫cos(3⁢x)d⁢x.

Worked solution

  1. 1

    Find the derivative pattern.

    dd⁢xsin(3⁢x)=3cos(3⁢x)

    The chain rule contributes the multiplier 3.

  2. 2

    Undo the extra multiplier.

    ∫cos(3⁢x)d⁢x=13sin(3⁢x)+C

    Dividing by 3 leaves the required cosine coefficient 1.

  3. 3

    Verify the coefficient.

    dd⁢x(sin(3⁢x)3+C)=13⋅3cos(3⁢x)=cos(3⁢x)

    The integration adjustment cancels the chain-rule multiplier.

Answer

sin(3⁢x)⁢/3+C

03 / YOUR TURN

Try it yourself.

Check your understanding

Find ∫2e2⁢xd⁢x.

Show answer & reasoning

e2⁢x+C

Because dd⁢xe2⁢x=2e2⁢x, the coefficient already matches the integrand: ∫2e2⁢xd⁢x=e2⁢x+C.

Keep practising

6 worksheet questions · 18 marks

Read the questions online
  1. Find ∫(8x3+ex)d⁢x.
    3 marks
  2. Find ∫xd⁢x for x>0.
    3 marks
  3. Find ∫x−2d⁢x on x>0.
    3 marks
  4. Find ∫4e−2⁢xd⁢x.
    3 marks
  5. Find ∫3sin(2⁢x)d⁢x.
    3 marks
  6. Find ∫(2cosx−sinx)d⁢x.
    3 marks
Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

  1. Find ∫(8x3+ex)d⁢x.
    3 marks

    Apply the power and natural-exponential rules.

    2x4+ex+C
  2. Find ∫xd⁢x for x>0.
    3 marks

    Rewrite as x12, then divide x32 by 32.

    23x32+C
  3. Find ∫x−2d⁢x on x>0.
    3 marks

    Raise the power to −1 and divide by −1.

    −x−1+C
  4. Find ∫4e−2⁢xd⁢x.
    3 marks

    Divide coefficient 4 by −2.

    −2e−2⁢x+C
  5. Find ∫3sin(2⁢x)d⁢x.
    3 marks

    Use the negative-cosine antiderivative and divide by 2.

    −32cos(2⁢x)+C
  6. Find ∫(2cosx−sinx)d⁢x.
    3 marks

    Integrate the two terms separately; check both signs.

    2sinx+cosx+C
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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Mathematical Methods integration and logarithmic antiderivatives; original learning and assessment material, with by-hand methods and explicitly described numerical models.

Official curriculum source ↗

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