→ Calculus → u-Substitution
Updated 15 Aug 2009

u-Substitution — Changing Variables in Integrals

Copyright © 2002–2017 by Stan Brown

Summary: Substitution is a hugely powerful technique in integration. Though the steps are similar for definite and indefinite integrals, there are two differences, and many students seem to have trouble keeping them straight. This page sorts them out in a convenient table, followed by a side-by-side example.

The Procedure

Just to keep things simple we’ll assume the original variable is x. Naturally the same steps will work for any variable of integration.

Indefinite Integrals Definite Integrals
1 Define u for your change of variables. (Usually u will be the inner function in a composite function.)
2 Differentiate u to find du, and solve for dx.
3 Substitute in the integrand and simplify.
4 (nothing to do) Use the substitution to change the limits of integration. Be careful not to reverse the order. Example: if u = 3−x² then definite integral from 0 to 4 becomes definite integral from 3 to minus 13.
5 If x still occurs anywhere in the integrand, take your definition of u from step 1, solve for x in terms of u, substitute in the integrand, and simplify.
6 Integrate.
7 Substitute back for u, so that your answer is in terms of x. Evaluate with u at the upper and lower new limits, and subtract.

An Example

Here’s a complete example, with indefinite and definite integrals shown in parallel.

Indefinite Integral
integral of x^5 (x^3-5)^4 dx
Definite Integral
integral from -1 to 1 of x^5 (x^3-5)^4 dx
1 u = x³−5 (inner function)
2 du = 3x² dx
dx = du / (3x²)
3 integral of 6 x^3 u^4 du integral from x=-1 to 1 of 6 x^3 u^4 du
After the substitution, u is the variable of integration, not x. But the limits have not yet been put in terms of u, and this must be shown.
4 (nothing to do) u = x³−5
x = −1 gives u = −6; x = 1 gives u = −4
integral from -6 to -4 of x^3 u^4 du
5 The integrand still contains x (in the form x³). Use the equation from step 1, u = x³−5, and solve for x³ = u+5.
integral of 6u^5 + 30u^4 du integral from -6 to -4 of 6u^5 + 30u^4 du
6 u6 + 6u5 + C
= (u+6)u5 + C
(u+6)u^5 from -6 to -4
(Factoring, though not strictly necessary, will make the next step easier.)
7 (x³−5+6)(x³−5)5 + C
= (x³+1)(x³−5)5 + C (Answer)
(−4+6)(−4)5 − (−6+6)(−6)5
= 2(−1024) − 0
= −2048 (Answer)
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