The Core Idea
U-substitution reverses the chain rule. When you see an integral that looks like f(g(x))·g'(x)dx, let u = g(x) and it simplifies to f(u)du. The trick is recognising the pattern.
∫ f(g(x)) · g'(x) dx = ∫ f(u) du where u = g(x)
The Decision Guide — How to Pick u
📋 Step-by-Step Decision Process
Rule 1Look for a function and its derivative both present in the integrand
Rule 2Set u = the inside function of a composite (what's inside sin, cos, eˣ, raised to a power)
Rule 3Set u = an expression whose derivative appears (or almost appears) as a factor
Rule 4If a polynomial is under a root or raised to a power, set u = that polynomial
The Key Question
After setting u = g(x) and computing du = g'(x)dx, can you rewrite the entire integral using only u and du — with no x remaining? If yes, you've chosen correctly.
Example 1 — Function and Derivative Present
📋 ∫ 2x·cos(x²) dx
Spotx² is inside cos, and 2x (its derivative) is the other factor
Setu = x², du = 2x dx
Rewrite∫ cos(u) du = sin(u) + C
Answersin(x²) + C
Example 2 — Off by a Constant
📋 ∫ x·e^(x²) dx
Spotx² inside eˣ, and x is present (off by factor of 2)
Setu = x², du = 2x dx → x dx = du/2
Rewrite∫ eᵘ · (du/2) = ½∫eᵘ du = ½eᵘ + C
Answer½e^(x²) + C
Example 3 — Polynomial Under a Root
📋 ∫ (3x² + 1)⁴ · 6x dx
Setu = 3x² + 1, du = 6x dx
Rewrite∫ u⁴ du = u⁵/5 + C
Answer(3x² + 1)⁵ / 5 + C
Example 4 — Trig Inside
📋 ∫ cos(x) · e^sin(x) dx
Spotsin(x) is inside eˣ, and cos(x) (its derivative) is the other factor
Setu = sin(x), du = cos(x) dx
Answere^sin(x) + C
Example 5 — Definite Integral (Change Bounds)
📋 ∫₀¹ 2x(x²+1)³ dx
Setu = x²+1, du = 2x dx
New boundsx=0: u=1, x=1: u=2
Integral∫₁² u³ du = [u⁴/4]₁² = 16/4 − 1/4 = 15/4
When NOT to Use U-Substitution
If no derivative of the inner function appears as a factor, u-substitution won't work directly. Try integration by parts for products like x·sin(x), x·eˣ, or x·ln(x). Try partial fractions for rational functions. Try trig substitution for expressions like √(a²−x²).
Frequently Asked Questions
What if I can't eliminate all the x's after substituting?
Either you chose the wrong u, or the integral requires a different technique. Try u = a different part of the integrand. If you still can't eliminate x, the integral probably needs integration by parts or another method.
How do I change limits of integration for definite integrals?
If x goes from a to b, substitute into u = g(x) to get the new bounds: u goes from g(a) to g(b). This is cleaner than substituting back at the end. Always check: lower bound when x = a, upper bound when x = b.
What if I'm off by a constant?
This is fine and common. If du = 3x dx but you only have x dx, multiply and divide by 3: x dx = (1/3) du. Pull the constant outside the integral.