U-Substitution — The Chain Rule in Reverse

What is U-Substitution?

U-substitution reverses the Chain Rule. When an integrand contains a function and (essentially) its derivative, substitute u for the inner function. The integral transforms into a simpler one in terms of u.

When to Use It

Look for: a composite function where the derivative of the inner function also appears (possibly with a constant factor). Triggers: ∫f(g(x))·g'(x)dx → let u=g(x), du=g'(x)dx → ∫f(u)du.

Step-by-Step Process

• Step 1: Choose u — usually the inner function or the 'complicated part'.
• Step 2: Compute du = (du/dx)dx.
• Step 3: Substitute — replace g(x) with u and g'(x)dx with du.
• Step 4: Integrate in terms of u.
• Step 5: Back-substitute — replace u with the original expression.

Examples

∫2x(x²+1)⁵ dx: Let u=x²+1, du=2x dx. Integral becomes ∫u⁵ du = u⁶/6+C = (x²+1)⁶/6+C.

∫cos(3x)dx: Let u=3x, du=3dx, so dx=du/3. Integral: (1/3)∫cos(u)du = sin(u)/3+C = sin(3x)/3+C.

∫x·e^(x²)dx: Let u=x², du=2x dx. Integral: (1/2)∫eᵘ du = eᵘ/2+C = e^(x²)/2+C.

U-Substitution for Definite Integrals

Two approaches: (1) change the bounds to u-values, or (2) back-substitute and use original bounds. Approach 1 is cleaner: if u=g(x), when x=a: u=g(a), when x=b: u=g(b). The new integral: ∫g(a)^g(b) f(u)du.

Why U-Substitution Works

The Chain Rule says d/dx[F(g(x))] = F'(g(x))·g'(x). Reading this backwards: ∫F'(g(x))·g'(x)dx = F(g(x)) + C. U-substitution is the formal mechanism for recognising and applying this reversal. By letting u = g(x) and du = g'(x)dx, the integral ∫F'(g(x))·g'(x)dx becomes ∫F'(u)du = F(u) + C = F(g(x)) + C.

The substitution is not just a notational trick — it genuinely transforms the variable of integration. When you write "let u = x² + 1", you are changing the integration variable from x to u. The du = 2x dx step ensures the transformation is mathematically valid (it is the change-of-variables formula for integrals).

Spotting the Pattern

The tell-tale sign of a u-substitution candidate: a function composed with another function, and the inner function's derivative (or a constant multiple of it) appears elsewhere in the integrand. Look for these patterns:

• ∫(something)ⁿ · (derivative of something)dx → let u = something
• ∫e^(something) · (derivative of something)dx → let u = something
• ∫sin(something) · (derivative of something)dx → let u = something
• ∫(derivative of something) / (something)dx → let u = something (gets ln)

When the Coefficient Is Off

Often the derivative of your chosen u appears in the integrand but with the wrong constant factor. Fix this by compensating algebraically. Example: ∫x·(x²+1)³ dx. Let u = x²+1, du = 2x dx. But the integrand has x dx, not 2x dx. Solution: x dx = du/2. The integral becomes ∫u³·(du/2) = (1/2)∫u³ du = u⁴/8 + C = (x²+1)⁴/8 + C. Multiplying and dividing by the missing constant is always valid and almost always the right fix.

Five Fully Worked Examples

U-Substitution for Definite Integrals — Changing Bounds

When applying u-substitution to a definite integral, you have two valid approaches. Approach 1 (preferred): Change the bounds to u-values immediately. If u = g(x), the new bounds are u = g(a) and u = g(b). Evaluate the integral entirely in u — no back-substitution needed. Approach 2: Complete the substitution, find the antiderivative in x (by back-substituting), then apply the original bounds. Both give the same answer. Approach 1 is faster for definite integrals because it avoids back-substitution.

When U-Substitution Fails

U-substitution fails when the integrand's remaining terms cannot be expressed purely in terms of u and du. Example: ∫x²·e^x dx. Letting u = x gives du = dx, but leaves ∫u²·eᵘ du — not simpler. Letting u = eˣ gives du = eˣ dx, but the x² has nothing to absorb the eˣ. This integral requires Integration by Parts, not u-substitution. The diagnostic: if after substituting you still have x in the integrand (not expressible in terms of u), u-substitution will not work — try a different technique.