U-Substitution: The Chain Rule in Reverse

U-substitution is the chain rule, run backward.

The chain rule says: to differentiate f(g(x)), multiply f'(g(x)) by g'(x).

U-substitution says: if you see f'(g(x)) · g'(x) in an integral, you can simplify by letting u = g(x).

This is the most common integration technique. Master it, and half of all integrals become straightforward.

The Pattern

Chain rule (forward): d/dx [f(g(x))] = f'(g(x)) · g'(x)

U-substitution (reverse): ∫ f'(g(x)) · g'(x) dx = f(g(x)) + C

Look for a function and its derivative together. The derivative is often hiding in plain sight, possibly multiplied by a constant.

The Mechanics

Step 1: Choose u = (inside function).

Step 2: Compute du/dx = g'(x), then write du = g'(x) dx.

Step 3: Substitute everything: replace g(x) with u, and g'(x) dx with du.

Step 4: Integrate in u.

Step 5: Substitute back: replace u with g(x).

Example 1: Basic Substitution

∫ 2x · cos(x²) dx

Spot it: x² is inside the cosine, and its derivative 2x is right there.

Let u = x². Then du = 2x dx.

Substitute: ∫ cos(u) du = sin(u) + C

Substitute back: sin(x²) + C

Check: d/dx[sin(x²)] = cos(x²) · 2x = 2x cos(x²) ✓

Example 2: Adjusting Constants

∫ x · e^(x²) dx

Let u = x². Then du = 2x dx, so x dx = du/2.

∫ x · e^(x²) dx = ∫ e^u · (1/2) du = (1/2) eᵘ + C = (1/2) e^(x²) + C

The factor of 2 is missing in the original integral, so we compensate with 1/2.

Example 3: Powers of Inner Functions

∫ (2x + 1)⁵ dx

Let u = 2x + 1. Then du = 2 dx, so dx = du/2.

∫ u⁵ · (1/2) du = (1/2) · u⁶/6 + C = u⁶/12 + C = (2x + 1)⁶/12 + C

Check: d/dx[(2x+1)⁶/12] = 6(2x+1)⁵ · 2 / 12 = (2x+1)⁵ ✓

Example 4: Trigonometric Integrals

∫ tan(x) dx = ∫ sin(x)/cos(x) dx

Let u = cos(x). Then du = -sin(x) dx, so sin(x) dx = -du.

∫ (-1/u) du = -ln|u| + C = -ln|cos(x)| + C

Or equivalently: ln|sec(x)| + C

Example 5: Square Roots

∫ x/√(1 + x²) dx

Let u = 1 + x². Then du = 2x dx, so x dx = du/2.

∫ (1/2) u^(-1/2) du = (1/2) · 2u^(1/2) + C = √u + C = √(1 + x²) + C

Recognizing the Pattern

Ask yourself: is there a composite function f(g(x)) whose derivative might give me this integrand?

If yes, the inside function g(x) is your u.

Common triggers:

• e^(something) → let u = something
• (something)ⁿ → let u = something
• sin(something), cos(something) → let u = something
• ln(something) appears with 1/x type terms → let u = the argument of ln
• √(something) → let u = something

The "something" is whatever is inside. Its derivative should appear (up to a constant) elsewhere in the integrand.

Substitution with Definite Integrals

For definite integrals, change the bounds when you substitute.

∫₀² x(x² + 1)³ dx

Let u = x² + 1. Then du = 2x dx, so x dx = du/2.

When x = 0: u = 1. When x = 2: u = 5.

∫₁⁵ (1/2) u³ du = (1/2) · [u⁴/4]₁⁵ = (1/8)[625 - 1] = 624/8 = 78

You never substitute back to x. Compute everything in u with new bounds.

When Substitution Doesn't Work

Not every integral yields to substitution.

∫ e^(x²) dx — no simple antiderivative exists.

∫ x² eˣ dx — needs integration by parts, not substitution.

∫ 1/(x² + 1)² dx — needs partial fractions or trig substitution.

Substitution works when the derivative of the inside function appears outside. When it doesn't, you need other tools.

Variations and Tricks

Completing the square: For ∫ 1/(x² + 2x + 5) dx, complete the square to get ∫ 1/((x+1)² + 4) dx, then substitute u = x + 1.

Algebraic manipulation: Sometimes multiply and divide by a constant, or add and subtract terms, to create the needed form.

Backwards thinking: If you know d/dx[tan(x)] = sec²(x), then ∫ sec²(x) dx = tan(x) + C. Every derivative formula gives you an integral formula for free.

Practice Recognizing Patterns

Train your eye to see the chain rule hiding:

∫ sec²(3x) dx — inside is 3x, derivative is 3, divide by 3 → (1/3)tan(3x) + C

∫ e^(5x) dx — inside is 5x, derivative is 5, divide by 5 → (1/5)e^(5x) + C

∫ cos(x)/sin(x) dx — inside is sin(x), derivative is cos(x) → ln|sin(x)| + C

∫ x² · e^(x³) dx — inside is x³, derivative is 3x², close enough → (1/3)e^(x³) + C

With practice, you'll do simple substitutions mentally.

Summary

U-substitution transforms integrals by replacing a composite function with a single variable.

1. Let u = inside function.
2. Express dx in terms of du.
3. Substitute and integrate.
4. Substitute back (indefinite) or change bounds (definite).

It's the chain rule in reverse. Every application of the chain rule in differentiation gives you a pattern for substitution in integration.

Further Reading

• Stewart, J. Calculus. Extensive substitution examples.
• Larson, R. Calculus. Step-by-step worked problems.
• Paul's Online Math Notes — Free detailed integration practice.

This is Part 5 of the Integrals series. Next: "Integration by Parts" — handling products of functions.

Part 5 of the Calculus Integrals series.

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