U-substitution is the integration counterpart of the chain rule. If the integrand has the form f(g(x)) * g'(x), substituting u = g(x), du = g'(x) dx transforms the integral into the simpler integral of f(u) du. This is the most commonly used integration technique. For definite integrals, you must also change the bounds from x-values to u-values.
Start by identifying the inner function u and checking that its derivative (or a constant multiple) appears in the integrand. Practice recognizing the pattern. Work through many examples with increasing complexity. Emphasize changing bounds for definite integrals (or converting back to x before evaluating).
U-substitution reverses the chain rule. When you differentiated f(g(x)) with the chain rule, the result was f'(g(x)) · g'(x). U-substitution works backward: when the integrand has that structure — an outer function applied to an inner function, multiplied by the inner function's derivative — you can "undo" the chain rule in one move.
The core technique: choose u = g(x) (the inner function), write du = g'(x) dx, and rewrite the entire integral in terms of u. If the substitution is correct, all the x's disappear and what remains is a simpler integral ∫f'(u) du. After integrating, substitute back to get the answer in terms of x.
Recognizing a good substitution is the real skill. For ∫ 2x·cos(x²) dx, notice that x² is the inner function and 2x — its derivative — already appears in the integrand. Setting u = x² gives du = 2x dx, transforming the integral to ∫ cos(u) du = sin(u) + C = sin(x²) + C. When the derivative is off by a constant (e.g., ∫ x·cos(x²) dx), you can compensate: du = 2x dx means x dx = du/2, so the integral becomes (1/2)∫ cos(u) du.
For definite integrals, the bounds must change. If you integrate ∫[0 to 1] with u = x², the new limits are u = 0² = 0 and u = 1² = 1. In general the limits become g(a) and g(b). A common error is keeping the original x-limits while integrating in u — this mixes two different variables and gives a wrong answer. Either change the bounds, or convert the antiderivative back to x before evaluating.
A failed substitution announces itself clearly: if you substitute and x-terms remain that can't be expressed in u, the choice was wrong. A good substitution leaves a purely u-based integral that is simpler than what you started with. With practice, spotting the "inner function whose derivative is present" becomes automatic — and u-substitution, along with integration by parts, will handle the majority of integrals you encounter.