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Integration by Substitution

Integration by substitution changes a difficult integral into a standard one by putting u=g(x), so that du=g'(x) dx matches part of the integrand.

In one line: Substitution is the chain rule walked backward.

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Understand it

Substitution is useful when the integrand contains a composite expression and its derivative, up to a constant factor. After integrating in u, the answer must be written back in terms of x for an indefinite integral.

Example

For ∫2x cos(x2) dx, take u=x2, so du=2x dx. The integral becomes ∫cos u du=sin u+C=sin(x2)+C.

Formula and rule

Method condition: if the integral has the form ∫f(g(x))g'(x) dx, put u=g(x), du=g'(x) dx, then integrate ∫f(u) du and substitute back.

For definite integrals, either change limits to u-values or substitute back before applying original limits.

See it worked out

Follow the example line by line, then use the same method on new questions.

Worked example

Evaluate ∫x√(x2+5) dx.

  1. 1Let u=x2+5, so du=2x dx and x dx=du/2.
  2. 2Integral =(1/2)∫u1/2 du=(1/2)·(2/3)u3/2+C=(1/3)(x2+5)3/2+C.

Method you can reuse

  1. 1Choose the inner expression as u.
  2. 2Differentiate it to find du.
  3. 3Rewrite all x-dependent parts and dx in terms of u.
  4. 4Integrate using a standard formula.
  5. 5Convert back to x for indefinite integrals.
  6. 6For definite integrals, handle limits consistently.

Visual tip

No diagram is required. The key representation is the chain rule in reverse: derivative of the inside expression must be accounted for.

Check yourself

Answer the question in your head before opening the answer.

Try it yourself

Evaluate ∫3(3x+1)4 dx.

Think first, then show answer
Let u=3x+1, du=3 dx. Integral =∫u4 du=u5/5+C=(3x+1)5/5+C.

Common wrong answer

Taking u=x2+5 in ∫x√(x2+5) dx but writing x dx=du instead of du/2, causing the answer to be twice the correct value.

Don't mix this up

A common error is choosing u correctly but forgetting to replace dx and the remaining factor, leaving a mixture of u and x.

Exam tip

Look for an inside function whose derivative is also present. If the derivative is present with a constant multiplier, substitution is likely efficient.

Write it in the exam

Each point is worth about one mark. Write point 1 for a 1-mark question and keep adding points as the marks go up.

How CBSE asks it

Appears in short-answer and long-answer questions with algebraic, exponential, logarithmic, and trigonometric composite functions. Definite integral versions may test correct transformation of limits.
  1. 1

    Enough for 1 mark

    Integration by substitution changes a difficult integral into a standard one by putting u=g(x), so that du=g'(x) dx matches part of the integrand.

  2. 2

    Add for 2 marks

    Example: For ∫2x cos(x2) dx, take u=x2, so du=2x dx. The integral becomes ∫cos u du=sin u+C=sin(x2)+C.

  3. 3

    Add for 3 marks

    Substitution is useful when the integrand contains a composite expression and its derivative, up to a constant factor.

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