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🌌 Calculus Kingdom · Calculus

Integration by Substitution

Undo the chain rule: choose u, adjust the constant, integrate powers of an inside function, trigonometric, exponential and logarithm composites and radicals, change the limits of a definite integral, and put x back in terms of u when one is left over.

In short

  • Substitution is the chain rule read backwards: name the inside u, and look for its derivative already present in the integrand.
  • Only a constant may be moved outside the integral. If replacing dx leaves a stray x behind, the substitution chosen was the wrong one.
  • In a definite integral the limits belong to x, so they must be rewritten in u — or the antiderivative must be written back in x before the original limits are used.
  • When a lone x survives the substitution, rearrange u = x + c into x = u - c and rewrite it; every term then integrates as an ordinary power.

The chain rule, run backwards

Differentiate (x2 + 1)4 and the chain rule hands you 4(x2 + 1)3 times 2x — the outer rule, then a copy of the derivative of the inside.

So if an integral already contains a copy of the derivative of something inside it, the chain rule can be run in reverse. That is the whole idea of substitution.

Look for two things together: an inside expression, and its derivative sitting somewhere in the integrand as a factor. In x(x2 + 1)3 the inside is x2 + 1, whose derivative is 2x, and an x is right there in front. In cos(3x) the inside is 3x, whose derivative is the constant 3.

Give that inside expression the name u. Everything else follows.

The recipe

Five steps, always the same:

  • Spot the inside. Which part is buried in a bracket, under a root, inside a function, or on the bottom of a fraction? Call it u.
  • Work out du. Differentiate u and attach dx: if u = x2 + 1 then du = 2x dx. The dx matters — it is the part being replaced.
  • Adjust the constant. If du asks for 2x and the integral offers only x, then x dx = du/2, and the 1/2 comes outside. Only a constant may cross the integral sign this way. If what is left over contains an x, the substitution was the wrong one.
  • Integrate in u. The whole point is that the new integral is one you already know: un, eu, sin(u), cos(u), 1/u.
  • Put x back. Replace u by what it stood for. Then check by differentiating: the chain rule should hand back the integrand you started with.

Because an antiderivative is only defined up to a constant, every answer really ends "+ C". In this skill the prompts say to omit it, so that two correct answers can be compared.

The five families

Almost every substitution you meet is one of these.

  • Powers. The integral of (3x + 2)4 dx uses u = 3x + 2, du = 3 dx: raise the exponent to 5, divide by 5, and divide by the 3 as well, giving (3x + 2)5/15. The integral of x(x2 + 4)3 dx uses u = x2 + 4 and x dx = du/2, giving (x2 + 4)4/8.
  • Trigonometric and exponential. The integral of cos(5x) dx is sin(5x)/5; the integral of sin(5x) dx is -cos(5x)/5 (the minus belongs to the rule, not to the substitution). The integral of x ex^2 dx uses u = x2 and comes out as ex^2/2.
  • Logarithms. Whenever the top of a fraction is the derivative of the bottom, the answer is a logarithm: the integral of 1/(2x + 5) dx is ln(2x + 5)/2 wherever 2x + 5 is positive, the integral of x/(x2 + 7) dx is ln(x2 + 7)/2, and the integral of ex/(ex + 1) dx is ln(ex + 1). The reverse power rule cannot do these, because raising the exponent -1 by one gives 0 and dividing by 0 is not allowed. That single gap is what the logarithm covers.
  • Radicals. Write the root as a power first. The integral of sqrt(2x + 1) dx is (2x + 1)3/2/3, the integral of 1/sqrt(2x + 1) dx is sqrt(2x + 1), and the integral of x/sqrt(x^2 + 9) dx is sqrt(x^2 + 9).
  • The logarithm as an inside. The integral of ln(x)/x dx looks like a logarithm problem and is not: u = ln(x) has du = dx/x, so it is the ordinary power rule on u, and the answer is ln(x)2/2.

Definite integrals: change the limits

A definite integral carries two limits, and those limits belong to x. Once you substitute, the variable is u, so the limits must be rewritten in u too — by putting each x into the substitution.

For the integral of x(x2 + 1)3 from x = 0 to x = 1: with u = x2 + 1, x = 0 gives u = 1 and x = 1 gives u = 2. The integral becomes (1/2) times the integral of u3 du from 1 to 2, which is (16 - 1)/8 = 15/8.

Changing the limits saves you from substituting back at all — you never have to write the answer in x. The alternative is to find the antiderivative in x first and then use the original limits; both are right, but mixing them is not. Putting the x limits into a u antiderivative measures something else entirely.

Some limits are chosen so the arithmetic stays exact: the integral of cos(3x) from 0 to pi/6 becomes the integral of cos(u) du/3 from 0 to pi/2, which is 1/3. And the integral of ln(x)/x from x = 1 to x = e becomes the integral of u du from 0 to 1, because ln(1) = 0 and ln(e) = 1.

When an x is left over

Sometimes du cleans up part of the integrand and leaves a lone x behind — the integral of x sqrt(x + 1) dx, for instance. Here u = x + 1 gives du = dx, so the root is handled, but the x in front is still an x.

The fix is to rearrange the substitution and put x back in terms of u: u = x + 1 means x = u - 1. Now the integrand is entirely in u:

  • (u - 1)sqrt(u) du = (u3/2 - u1/2) du, two ordinary powers.

Integrating gives (2/5)u5/2 - (2/3)u3/2, and substituting back gives 2(x + 1)5/2/5 - 2(x + 1)3/2/3.

The same move works for the integral of x/(x + 2)2 dx: x = u - 2 turns it into (1/u - 2/u2) du, whose antiderivative is ln(u) + 2/u, that is ln(x + 2) + 2/(x + 2). The two minus signs there — one from the split, one from integrating u-2 — cancel, which is a favourite place to lose a sign.

Worked examples

Example 1

Find the antiderivative of f(x) = (4x - 3)5. Omit the "+ C".

  1. The inside expression is 4x - 3, so let u = 4x - 3.
  2. Differentiate: du = 4 dx, so dx = du/4.
  3. The integral becomes (1/4) times the integral of u5 du.
  4. Integrate: (1/4)(u6/6) = u6/24.
  5. Substitute back: (4x - 3)6/24.
  6. Check: differentiating gives 6(4x - 3)5(4)/24 = (4x - 3)5, the original function.

Example 2

Find the antiderivative of f(x) = 6x/(x2 + 5). Omit the "+ C".

  1. The bottom is x2 + 5 and its derivative is 2x, which is the top up to a constant. Let u = x2 + 5.
  2. Then du = 2x dx, so x dx = du/2.
  3. The integral becomes (6/2) times the integral of du/u = 3 times the integral of du/u.
  4. The integral of du/u is ln(u), so the result is 3ln(u).
  5. Substitute back: 3ln(x2 + 5).
  6. No absolute value is needed here, because x2 + 5 is positive for every x.

Example 3

Evaluate the definite integral of x sqrt(x^2 + 9) from x = 0 to x = 4. Give an exact answer.

  1. Let u = x2 + 9, so du = 2x dx and x dx = du/2.
  2. Change the limits: x = 0 gives u = 9, and x = 4 gives u = 25.
  3. The integral becomes (1/2) times the integral of sqrt(u) du from 9 to 25.
  4. An antiderivative of sqrt(u) is (2/3)u3/2, and half of that is (1/3)u3/2.
  5. u3/2 is the cube of the square root: at u = 25 it is 53 = 125, and at u = 9 it is 33 = 27.
  6. Value = (125 - 27)/3 = 98/3.

Example 4

Find the antiderivative of f(x) = x sqrt(x + 4), for x where x + 4 > 0. Omit the "+ C".

  1. Let u = x + 4, so du = dx. The root becomes sqrt(u), but the x in front is still an x.
  2. Rearrange the substitution: x = u - 4.
  3. The integral becomes the integral of (u - 4)u1/2 du = the integral of (u3/2 - 4u1/2) du.
  4. Integrate each power: (2/5)u5/2 - 4(2/3)u3/2 = (2/5)u5/2 - (8/3)u3/2.
  5. Substitute back: 2(x + 4)5/2/5 - 8(x + 4)3/2/3.

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

Which substitution turns the integral of cos(3x) dx into an integral you can do straight away?

  1. u = 3x
  2. u = sin(3x)
  3. u = cos(3x)
  4. u = x

Answer: A. u = 3x

  1. The inside expression is 3x.
  2. Its derivative gives du = 3 dx, which the integral already offers up to the constant 3.
  3. So the substitution is u = 3x.

Problem 2

Difficulty 3 of 5

Find the antiderivative of f(x) = x(x2 + 7)2. Give an expression in x and OMIT the "+ C".

Answer: (x2 + 7)3/6

  1. Let u = x2 + 7, so du = 2x dx and x dx = du/2.
  2. The integral becomes (1/2) times the integral of u2 du.
  3. Integrating: (1/2)(u3/3) = u3/6.
  4. Substituting back: (x2 + 7)3/6.

Problem 3

Difficulty 4 of 5

Find the antiderivative of f(x) = x sin(4x2), with x in radians. Give an expression in x and OMIT the "+ C".

Answer: -cos(4x2)/8

  1. Let u = 4x2, so du = 8x dx and x dx = du/8.
  2. The integral becomes (1/8) times the integral of sin(u) du.
  3. That integrates to -cos(u)/8.
  4. Substituting back: -cos(4x2)/8.

Common mistakes

  • Forgetting the constant that du leaves outside. The integral of cos(3x) dx is sin(3x)/3, not sin(3x): differentiating your answer shows the spare 3 at once.
  • Keeping the factor that du already used up, writing x(x2 + 1)4/8 instead of (x2 + 1)4/8. The x became part of du and cannot appear twice.
  • Using the old x limits on a u antiderivative. After substituting, the limits have to be changed too, or the answer measures a different region.
  • Reaching for the power rule on 1/(2x + 5). An exponent of -1 is the one case the reverse power rule cannot handle, and the logarithm is what covers it.
  • Losing a sign in the back-substitution: u = x + 3 means x = u - 3, so the second term is subtracted. In x/(x + 3)2 that minus meets the minus from integrating u-2 and the two cancel.

What you should be able to do

  • Choose the substitution u, find du, and work out the constant factor it leaves outside.
  • Integrate (ax + b)n, x(x2 + c)n, cos(kx), x ex^2, sqrt(ax + b) and similar composites.
  • Recognise the forms that integrate to a logarithm.
  • Evaluate a definite integral by changing the limits along with the variable, and rearrange u = x + c into x = u - c when a lone x survives.

Where this fits in the curriculum

Common Core

  • FUN-6.C

    AP Calculus AB, Unit 6 — Determine antiderivatives of functions and indefinite integrals, using knowledge of derivatives.

    The Common Core has no calculus standards; this is the AP Calculus AB learning objective.

  • FUN-6.D

    AP Calculus AB, Unit 6 — For integrands requiring substitution or rearrangement into equivalent forms, determine indefinite integrals and evaluate definite integrals.

    The Common Core has no calculus standards; this is the AP Calculus AB learning objective.

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