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Year 12 · Integration techniques and applications · 22 min

Substitution

Choose substitutions that turn a composite integrand into a simpler variable, including definite integrals.

01 / LEARN

The idea, step by step

Reverse a chain rule

Substitution works when an inner expression and a multiple of its derivative occur together. Let u equal the inner expression, calculate du, and replace every remaining x term before integrating. The substitution is an equality of differentials, not permission to ignore a coefficient.

For ∫x⁢(1+x2)4d⁢x, u=1+x squared gives du=2x dx, so x dx=du/2. The transformed integral is a power of u. A derivative check in x confirms the factor.

Keep variables and limits consistent

For a definite integral, change the endpoints into u values and finish in u. Alternatively, return to x before using the original endpoints. Do not use x endpoints in an antiderivative still written in u.

A helpful substitution must simplify the whole integrand. If an unexplained x remains after substitution, either express it in u using the domain information or choose another method. State any restrictions needed for the substitution.

A common trap

Change the variable but keep the original endpoints without converting back.

Instead: Limits and antiderivative must refer to the same variable.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example 1

Find ∫x⁢(1+x2)4d⁢x.

  1. Set u=1+x squared, du=2x dx.
  2. Integrate 12u4 and substitute back.
(1+x2)5/10+C.
Example 2

Worked example 2

Evaluate ∫012⁢xex2d⁢x.

  1. Use u=x squared. The new limits are 0 and 1.
  2. Integrate eu between those limits.
e−1.

Check your understanding

For u=3x+1, what is dx?

Show answer

d⁢u⁢/3.

Since du=3dx, divide by 3.

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Read the questions online
  1. Find ∫2⁢x⁢/⁢(x2+1)d⁢x.
    2 marks
  2. Find ∫cosxsin3xd⁢x.
    2 marks
  3. Find ∫x⁢/4+x2d⁢x.
    2 marks
  4. Find ∫(3⁢x+1)5d⁢x.
    2 marks
  5. Find ∫e2⁢x/⁢(1+e2⁢x)d⁢x.
    2 marks
  6. Evaluate ∫02x⁢(1+x2)d⁢x.
    2 marks
Read the worked solutions

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

  1. Find ∫2⁢x⁢/⁢(x2+1)d⁢x.
    2 marks

    Use u=x squared+1.

    The integral becomes ∫d⁢u⁢/⁢u; u is positive.

    ln(x2+1)+C.
  2. Find ∫cosxsin3xd⁢x.
    2 marks

    Set u=sin x and du=cos x dx.

    Integrate u cubed.

    sin4x⁢/4+C.
  3. Find ∫x⁢/4+x2d⁢x.
    2 marks

    Use u=4+x squared and x dx=du/2.

    Integrate u−1⁢/2/2.

    4+x2+C.
  4. Find ∫(3⁢x+1)5d⁢x.
    2 marks

    Use u=3x+1, dx=du/3.

    Integrate u to the fifth and divide by 3.

    (3⁢x+1)6/18+C.
  5. Find ∫e2⁢x/⁢(1+e2⁢x)d⁢x.
    2 marks

    Use u=1+exp(2x); du=2exp(2x)dx.

    Integrate du/(2u).

    12ln(1+e2⁢x)+C.
  6. Evaluate ∫02x⁢(1+x2)d⁢x.
    2 marks

    With u=1+x squared, the limits become 1 and 5.

    12∫15ud⁢u=(25−1)⁢/4=6.

    6.
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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics integration techniques and applications, including inverse-trigonometric patterns, substitution, parts, simple partial fractions, areas and solids of revolution.

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