U-Substitution

The formula and its origin

The substitution rule:

∫ f(g(x)) · g'(x) dx  =  ∫ f(u) du,   where u = g(x), du = g'(x) dx.

It's the chain rule with the equality run backwards. If F is an antiderivative of f, then by the chain rule:

d/dx [ F(g(x)) ]  =  F'(g(x)) · g'(x)  =  f(g(x)) · g'(x).

Integrating both sides gives ∫f(g(x))·g'(x)dx = F(g(x)) + C — and the right-hand side is exactly what you'd get by writing F(u) + C and then substituting u = g(x). The substitution rule is the chain rule, rewritten in integral notation.

How to apply it

The mechanical recipe:

1. Scan the integrand for a composition f(g(x)) and a factor close to g'(x).
2. Set u = g(x).
3. Compute du = g'(x) dx.
4. Rewrite the integral entirely in u — substitute g(x) → u and g'(x) dx → du.
5. Integrate ∫f(u)du.
6. Substitute back u = g(x) to express the answer in x.
7. For definite integrals — either change the limits to u(a), u(b) OR substitute back to x before evaluating. Pick one, stay consistent.

The hard step is step 1. The rest is bookkeeping.

Worked example — ∫ 2x · sin(x²) dx

The inner function is x²; its derivative is 2x, which appears as a factor. Perfect candidate.

u  = x²
du = 2x dx

∫ 2x · sin(x²) dx  =  ∫ sin(u) du  =  −cos(u) + C
                                    =  −cos(x²) + C.

Verify by differentiating — d/dx[−cos(x²) + C] = sin(x²)·2x. Matches the integrand exactly. The composition disappeared; we integrated a flat ∫sin(u)du and put the layers back at the end.

Worked example — ∫ sin(3x) dx (constant rescale)

The inner function is 3x; its derivative is 3, which doesn't appear in the integrand. Set u = 3x anyway; rescale du to fit.

u  = 3x
du = 3 dx   ⟹  dx = du/3

∫ sin(3x) dx  =  ∫ sin(u) · (du/3)
              =  (1/3) · ∫ sin(u) du
              =  −(1/3) cos(u) + C
              =  −(1/3) cos(3x) + C.

The factor of 3 was missing, but constants flow naturally — divide by 3 outside the integral. This works any time the missing factor is a constant.

Worked example — ∫₀¹ x · e^(x²) dx (definite, change limits)

u  = x²
du = 2x dx  ⟹  x dx = du/2

When x = 0,  u = 0.
When x = 1,  u = 1.

∫₀¹ x · e^(x²) dx  =  ∫₀¹ e^u · (du/2)
                    =  (1/2) [e^u]₀¹
                    =  (1/2) (e − 1).

Notice the limits changed from x = 0, 1 to u = 0, 1 (coincidentally the same here, but in general u(a) ≠ a). The cleaner alternative is to leave the limits in x and substitute back to x at the very end — both are correct, but mixing them is a common error.

Worked example — ∫ tan(x) dx

Rewrite tan(x) = sin(x)/cos(x). Spot the structure — cos(x) inside, and sin(x) is (up to a sign) the derivative of cos(x).

u  = cos(x)
du = −sin(x) dx   ⟹  sin(x) dx = −du

∫ tan(x) dx  =  ∫ sin(x)/cos(x) dx
             =  ∫ (−du)/u
             =  −∫ (1/u) du
             =  −ln|u| + C
             =  −ln|cos(x)| + C
             =  ln|sec(x)| + C.

The negative sign comes from the −sin in du. Many tan-integration mistakes drop this sign. Verify — d/dx[ln|sec(x)|] = (sec(x)·tan(x))/sec(x) = tan(x). Matches.

U-substitution vs other techniques

	U-substitution	Integration by parts	Trig substitution	Partial fractions
Best when	Inner function + its derivative present	Product of dissimilar functions (no derivative chain)	√(a²±x²) or √(x²−a²) in integrand	Rational function P(x)/Q(x)
Underlying rule	Chain rule (reverse)	Product rule (reverse)	Pythagorean identity	Algebraic decomposition
Order to try	First — fastest	Second — when u-sub fails	When square roots resist u-sub	For rational functions specifically
Mechanical?	Very — once pattern is spotted	Less so — LIATE helps	Mechanical with the three triangle templates	Mechanical with Heaviside cover-up
Speed	Fast	Medium	Slow (triangle setup)	Slow (algebra-heavy)
Hardest part	Spotting the composition	Choosing u and dv	Picking the right trig sub	Factoring Q(x)

When u-substitution is the right tool

• Polynomial × derivative. ∫xⁿ·polynomial(x^(n+1))dx is always u-sub. Any time a polynomial-degree-n factor multiplies a function of polynomial-degree-(n+1) inside.
• Exponentials with linear arguments. ∫e^(ax+b)dx, ∫a^x dx, ∫sin(kx)dx — all reduce by u = (linear).
• Trig and inverse trig. ∫tan(x)dx, ∫cot(x)dx, ∫sec(x)tan(x)dx, ∫(1/√(1−x²))dx — all yield to u-sub on the appropriate inner.
• Square roots with their derivatives. ∫x·√(x²+1)dx = u = x²+1.
• Logarithm and derivative paired. ∫(ln x)ⁿ/x dx with u = ln x.
• Probability density transforms. Change-of-variable formula for probability density functions is the multivariable form of u-substitution.

Common mistakes

• Forgetting du. Rewriting the integrand without converting g'(x)dx to du gives garbage. Every dx must become du (or du/constant).
• Mixing limits and variables. If you change limits to u-values, your integrand must be in u when you evaluate. Don't substitute back to x and then plug in u-limits.
• Dropping the constant rescale. ∫sin(3x)dx with u = 3x gives dx = du/3 — that 1/3 must appear. Skipping it is the most common arithmetic error.
• Sign errors from negative du. u = cos(x) gives du = −sin(x)dx. The negative sign carries through.
• Substituting incomplete expressions. If only part of the integrand is in terms of g(x), the substitution is incomplete — you can't replace everything with u, du.
• Forgetting + C. Indefinite integrals always need the constant of integration. Standard convention; lose it and lose the point on a graded answer.

Connections to other techniques

Chain rule. U-substitution is the chain rule, literally. If you can read d/dx[F(g(x))] = F'(g(x))·g'(x) right-to-left, you can do u-sub. The mental motion is identical.

Integration by parts. When u-sub fails because there's no inner-with-its-derivative pairing, parts is the next tool. ∫x·sin(x)dx isn't u-sub (sin(x) has no x as derivative), it's parts.

Trig substitution. A specialized u-sub where u is a trig function chosen to exploit Pythagorean identities. Same idea, different recognition pattern.

Change of variable in multivariable integrals. The Jacobian-determinant change-of-variable formula ∫∫_R f(x,y) dA = ∫∫_S f(x(u,v), y(u,v))|J|du dv is the multivariable generalization. Same theorem, more bookkeeping.

Numerical integration. When a closed-form antiderivative exists via u-sub, the resulting symbolic answer is more accurate than any numerical quadrature. U-sub is also a preprocessing step for adaptive quadrature on integrands with sharp inner-function features.