When to Use U-Substitution
U-substitution is the integration counterpart of the chain rule for derivatives. Whenever you see an integral where part of the integrand is a composite function — a function inside another function — and the derivative of the inner function also appears in the integrand, u-substitution is likely the right technique. The classic signal is an integral of the form ∫f(g(x)) · g'(x) dx.
For instance, in ∫2x · cos(x²) dx, you can spot that x² sits inside cosine and its derivative 2x is right there as a factor. That is a perfect setup for u-sub.
More generally, look for an integrand that contains a "chunk" — an expression nested inside a power, a trig function, an exponential, a logarithm, or a square root — and check whether the derivative of that chunk (or a constant multiple of it) appears as a separate factor. If it does, u-substitution will simplify the integral dramatically.
If the derivative of the inner piece is nowhere to be found, u-sub probably will not help and you should consider other techniques like integration by parts or partial fractions.
The Method Step by Step
Here is the systematic procedure for u-substitution.
- Step 1 — Choose u.
Identify the inner function (the "chunk" inside another function) and set
u = g(x). - Step 2 — Find du.
Differentiate u with respect to x to get
du = g'(x) dx. - Step 3 — Rewrite the integral.
Replace every occurrence of g(x) with u and replace
g'(x) dxwith du. The goal is an integral written entirely in terms of u with no remaining x's. - Step 4 — Integrate.
Evaluate the simpler integral in u.
- Step 5 — Substitute back.
Replace u with the original expression g(x) to get the final answer in terms of x, and add the constant
C.
As a concrete example, evaluate ∫2x · cos(x²) dx.
Let u = x², so du = 2x dx.
The integral becomes ∫cos(u) du = sin(u) + C = sin(x²) + C.
sin(x²) + CEvery u-substitution follows this exact five-step pattern, no matter how complex the integrand looks.
Choosing the Right u
Selecting the right u is the hardest part of u-substitution, and it gets easier with practice.
A reliable rule of thumb: let u be the inner function — the expression that is nested inside a power, trig function, exponential, square root, or logarithm.
In ∫(3x + 1)⁵ dx, the inner piece is 3x + 1, so let u = 3x + 1.
In ∫e^(4x) dx, the exponent 4x is the inner function, so u = 4x.
In ∫sin(x) cos(x) dx, you can let u = sin(x) (then du = cos(x) dx) or u = cos(x) (then du = −sin(x) dx) — both work.
Try the denominator for fractions
Another helpful guideline: if the integrand contains a fraction, try letting u equal the denominator. For example, in ∫x/(x² + 1) dx, setting u = x² + 1 gives du = 2x dx, and you can account for the factor of 2 by writing ½ du = x dx.
When your substitution leaves behind x-terms that cannot be expressed in terms of u, your choice is wrong — go back and try a different u. Constant multiples are fine to adjust (factor them out front), but leftover x's are a sign to rethink your approach.
Definite Integrals with U-Sub
When applying u-substitution to a definite integral, you have two options — and choosing wisely saves time.
| Option | Approach | Worked on ∫₀² 2x(x² + 1)³ dx | Result |
|---|---|---|---|
| 1 — Change the bounds | When you substitute u = g(x), convert the limits of integration from x-values to u-values. You never substitute back to x because the bounds are already in terms of u. | If u = x² + 1, then when x = 0, u = 1, and when x = 2, u = 5. The integral becomes ∫₁⁵ u³ du = [u⁴/4]₁⁵ = (625/4) − (1/4) = 624/4. | 156 |
| 2 — Substitute back, then evaluate | Find the antiderivative in terms of u, replace u with g(x), and then plug in the original x-limits. | Antiderivative is (x² + 1)⁴ / 4, evaluated from 0 to 2 gives (5⁴ − 1⁴)/4 = 624/4. | 156 |
Option 1 is generally faster and reduces the chance of errors, because you avoid the back-substitution step entirely.
Whichever method you use, never mix x-bounds with a u-expression or u-bounds with an x-expression — that is a guaranteed wrong answer.
Worked Examples
∫(2x + 3)⁴ dx.
Let u = 2x + 3, du = 2 dx, so dx = du/2.
Integral becomes (1/2)∫u⁴ du = (1/2)(u⁵/5) + C.
(2x + 3)⁵/10 + C∫x√(x² + 9) dx.
Let u = x² + 9, du = 2x dx, so x dx = du/2.
Integral becomes (1/2)∫√u du = (1/2)(2u^(3/2)/3) + C.
(x² + 9)^(3/2)/3 + C∫sin³(x)cos(x) dx.
Let u = sin(x), du = cos(x) dx.
Integral becomes ∫u³ du = u⁴/4 + C.
sin⁴(x)/4 + C∫e^(√x) / √x dx.
Let u = √x, du = 1/(2√x) dx, so dx/√x = 2 du.
Integral becomes 2∫eᵘ du = 2eᵘ + C.
2e^(√x) + C∫tan(x) dx = ∫sin(x)/cos(x) dx.
Let u = cos(x), du = −sin(x) dx.
Integral becomes −∫(1/u) du = −ln|u| + C.
−ln|cos(x)| + CEach of these follows the same five-step pattern — the only variation is in identifying the right substitution.
Common Mistakes
Even students who understand the concept of u-substitution make predictable errors.
After choosing u, you must express dx in terms of du. Writing ∫u³ dx instead of ∫u³ (du/2) is incorrect — every part of the integrand must be converted.
After substitution, no x-variable should remain. If it does, either solve for x in terms of u and substitute, or reconsider your choice of u.
For indefinite integrals, your final answer must be in terms of x, not u. Writing u⁵/10 + C as your answer is incomplete.
If du = 3 dx but the integrand only has dx, you need to write dx = du/3 and include the factor of 1/3. Dropping this constant is a common source of wrong answers.
If you change the limits to u-values, evaluate entirely in u. If you substitute back to x, use the original x-limits. Mixing the two produces nonsense.
If you want to double-check your substitutions and catch these errors before they cost you exam points, scan your work with Solver AI to see each step verified instantly.