Integration · Substitution Rule

The Substitution Rule — Reversing the Chain Rule

The chain rule tells us how to differentiate composite functions. U-substitution runs that process in reverse: by introducing a new variable u = g(x), we can transform a complicated integral into a simpler standard form and integrate it with ease.

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§ 01Derivation — The Chain Rule in Reverse

Every integration technique is a differentiation rule run backwards. U-substitution is the reverse chain rule, and it handles the largest class of non-trivial integrals you will encounter.

Recall the chain rule: if F is an antiderivative of f, then d/dx[F(g(x))] = f(g(x))·g′(x). Integrating both sides:

Chain Rule Integrated
∫ f(g(x)) · g′(x) dx = F(g(x)) + C

If we let u = g(x), then du = g′(x) dx, and the integral becomes ∫f(u) du, a standard integral in u.

Substitution Rule
∫ f(g(x)) · g′(x) dx = ∫ f(u) du
where u = g(x) and du = g′(x) dx
Prerequisites Fluent with differentiation (chain rule especially), basic antiderivative formulas, and algebraic manipulation. You should know: ∫xⁿ dx, ∫eˣ dx, ∫sin x dx, ∫cos x dx, ∫(1/x) dx.

§ 02How to Spot and Apply U-Substitution

Substitution works when the integrand contains a composite function f(g(x)) and the derivative g′(x) is also present (possibly as a scalar multiple).

The 5-Step U-Substitution Method

  1. Identify u. Choose u = g(x) to be the "inner function": typically the expression inside a power, root, exponential, or trig function.
  2. Compute du. Differentiate: du = g′(x) dx. Solve for dx if needed: dx = du/g′(x).
  3. Substitute. Replace every occurrence of g(x) with u and g′(x) dx with du. The integral should now be entirely in u with no x remaining.
  4. Integrate in u. Evaluate ∫f(u) du using standard antiderivatives.
  5. Back-substitute. Replace u with g(x) to express the answer in terms of x. Add +C.
Spotting the Pattern Look for a function and its derivative in the integrand. In ∫2x·cos(x²)dx, u = x² works because its derivative 2x is right there. In ∫x³√(x⁴+1)dx, u = x⁴+1 works because du = 4x³dx and x³ is present.

When a Constant Factor Is Missing

The derivative g′(x) might be present up to a constant multiple. We can always adjust by multiplying and dividing by the missing constant.

Adjusting for a Constant
∫ x sin(x²) dx: u = x², du = 2x dx, so x dx = du/2

∫x sin(x²) dx = ½∫sin(u) du = −½cos(u) + C = −½cos(x²) + C

§ 03Worked Examples — Indefinite Integrals

Example 1∫ 2x(x²+1)⁵ dx
  1. u = x²+1, du = 2x dx. (The factor 2x is present exactly.)
  2. ∫(x²+1)⁵·2x dx = ∫u⁵ du = u⁶/6 + C.
  3. Back-sub: (x²+1)⁶/6 + C.
∫2x(x²+1)⁵ dx = (x²+1)⁶/6 + C
Example 2∫ cos(3x) dx
  1. u = 3x, du = 3 dx, so dx = du/3.
  2. ∫cos(u)·(du/3) = (1/3)sin(u) + C.
  3. Back-sub: (1/3)sin(3x) + C.
∫cos(3x) dx = (1/3)sin(3x) + C
Example 3∫ x·e^(x²) dx
  1. u = x², du = 2x dx, so x dx = du/2.
  2. ∫e^u · du/2 = (1/2)eᵘ + C.
  3. Back-sub: (1/2)e^(x²) + C.
∫xe^(x²) dx = ½e^(x²) + C
Example 4∫ tan x dx = ∫ sin x / cos x dx
  1. u = cos x, du = −sin x dx, so sin x dx = −du.
  2. ∫(1/u)·(−du) = −ln|u| + C.
  3. Back-sub: −ln|cos x| + C = ln|sec x| + C.
∫tan x dx = −ln|cos x| + C = ln|sec x| + C
Example 5∫ x²√(x³+4) dx
  1. u = x³+4, du = 3x² dx, so x² dx = du/3.
  2. ∫√u · (du/3) = (1/3)·(⅔)u^(3/2) + C = (2/9)u^(3/2) + C.
  3. Back-sub: (2/9)(x³+4)^(3/2) + C.
∫x²√(x³+4) dx = (2/9)(x³+4)^(3/2) + C
Example 6∫ e^x / (1 + e^x) dx
  1. u = 1+eˣ, du = eˣ dx.
  2. ∫(1/u) du = ln|u| + C.
  3. Back-sub: ln(1+eˣ) + C.
∫eˣ/(1+eˣ) dx = ln(1+eˣ) + C
Example 7∫ sin³x cos x dx
  1. u = sin x, du = cos x dx. (cos x is present as the outer factor.)
  2. ∫u³ du = u⁴/4 + C.
  3. Back-sub: sin⁴x / 4 + C.
∫sin³x cos x dx = sin⁴x / 4 + C
Example 8∫ 1/(x ln x) dx
  1. u = ln x, du = (1/x) dx.
  2. ∫(1/u) du = ln|u| + C.
  3. Back-sub: ln|ln x| + C.
∫ 1/(x ln x) dx = ln|ln x| + C
TrickyExample 9 — ∫ x/(√(1−x²)) dx
  1. u = 1−x², du = −2x dx, so x dx = −du/2.
  2. ∫(1/√u)·(−du/2) = −½∫u^(−½) du = −½·2√u = −√u + C.
  3. Back-sub: −√(1−x²) + C.
∫x/√(1−x²) dx = −√(1−x²) + C

§ 04Substitution in Definite Integrals

For definite integrals, you have two approaches: (1) change the limits of integration to match u, or (2) back-substitute and use original limits. Method (1) is neater.

Definite Substitution — Change the Limits
∫ₐᵇ f(g(x))g′(x)dx = ∫_{g(a)}^{g(b)} f(u) du

When x = a, u = g(a). When x = b, u = g(b). You never need to back-substitute.

DefiniteExample 10 — ∫₀¹ 2x(x²+1)⁵ dx
  1. u = x²+1, du = 2x dx.
  2. Limits: x=0 → u=1; x=1 → u=2.
  3. ∫₁² u⁵ du = [u⁶/6]₁² = 64/6 − 1/6 = 63/6 = 21/2.
∫₀¹ 2x(x²+1)⁵ dx = 21/2 = 10.5
DefiniteExample 11 — ∫₀^(π/4) tan x dx
  1. From Example 4, antiderivative is −ln|cos x|.
  2. [−ln|cos x|]₀^(π/4) = −ln(1/√2) − (−ln 1) = −ln(1/√2) = ½ln 2.
∫₀^(π/4) tan x dx = ½ ln 2 ≈ 0.347
DefiniteExample 12 — ∫₀² x e^(x²) dx
  1. u = x², du = 2x dx, so x dx = du/2.
  2. Limits: x=0 → u=0; x=2 → u=4.
  3. ∫₀⁴ eᵘ · (du/2) = ½[eᵘ]₀⁴ = ½(e⁴−1).
∫₀² xe^(x²) dx = ½(e⁴−1) ≈ 26.80

§ 05Quick Reference — Common Substitutions

Integrand PatternChoose u =du =Result Form
∫[g(x)]ⁿ g′(x) dxg(x)g′(x) dxuⁿ⁺¹/(n+1) + C
∫eᵍ⁽ˣ⁾ g′(x) dxg(x)g′(x) dxeᵘ + C
∫g′(x)/g(x) dxg(x)g′(x) dxln|u| + C
∫sin(g(x))g′(x) dxg(x)g′(x) dx−cos u + C
∫cos(g(x))g′(x) dxg(x)g′(x) dxsin u + C
∫sin x / cos x dxcos x−sin x dx−ln|u| + C
∫f′(x)/(1+[f(x)]²) dxf(x)f′(x) dxarctan u + C

Common Mistakes

Mistake 1 — Not fully eliminating xAfter substituting u = g(x), every x in the integral must be replaced. If x still appears, the substitution is incomplete — try a different u, or solve du = g′(x)dx for x and substitute.
Mistake 2 — Forgetting to change limits in definite integralsIf you substitute u = g(x) in a definite integral and keep the original x-limits, you will compute the wrong value. Always convert limits to u-values: u_lower = g(a), u_upper = g(b).
Mistake 3 — Dropping the constant factorIf du = 5x² dx but the integrand has x² dx, you need to write x² dx = du/5 and include the 1/5 factor. Forgetting this multiplier is very common.
Mistake 4 — Using substitution when it isn't needed∫x²dx does not need substitution — it integrates directly to x³/3. Over-applying substitution wastes time. Ask: is there already a composite structure here?

§ 06Test Yourself — 10-Question Quiz

Score: 0 / 0
Q1 of 10

For ∫2x(x²+1)⁵ dx, the best substitution is u = ?

u = x²+1 because its derivative 2x is present. This turns the integral into ∫u⁵ du, which is straightforward.

Q2 of 10

Evaluate ∫cos(5x) dx. The answer has the form (1/k)sin(5x) + C. What is k?

  1. u = 5x, du = 5dx, dx = du/5.
  2. ∫cos(u)(du/5) = (1/5)sin u + C = (1/5)sin(5x) + C. So k = 5.
Q3 of 10

What is ∫tan x dx?

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

Q4 of 10

Evaluate ∫₀¹ 2x(x²+1)⁵ dx. Enter your answer.

  1. u=x²+1. New limits: u(0)=1, u(1)=2.
  2. ∫₁² u⁵ du = [u⁶/6]₁² = 64/6−1/6 = 63/6 = 10.5.
Q5 of 10

For ∫x e^(x²) dx, after substituting u = x², the integral becomes:

u = x², du = 2x dx, so x dx = du/2. The integral becomes ∫eᵘ (du/2) = ½∫eᵘ du.

Q6 of 10

What is ∫sin³x cos x dx?

u = sin x, du = cos x dx. ∫u³ du = u⁴/4 + C = sin⁴x/4 + C.

Q7 of 10

Evaluate ∫eˣ/(1+eˣ) dx. The answer is ln(?) + C.

u = 1+eˣ, du = eˣ dx. ∫(1/u) du = ln|u| + C = ln(1+eˣ) + C (no absolute value needed since 1+eˣ > 0).

Q8 of 10

Evaluate ∫₀² xe^(x²) dx. Enter your answer to 2 decimal places.

  1. u=x², du=2xdx. Limits: 0→0, 2→4.
  2. ½∫₀⁴ eᵘ du = ½[eᵘ]₀⁴ = ½(e⁴−1) ≈ ½(54.598−1) ≈ 26.80.
Q9 of 10

What is ∫1/(x ln x) dx?

u = ln x, du = (1/x) dx. ∫(1/u) du = ln|u| + C = ln|ln x| + C.

Q10 of 10

For a definite integral with substitution, when should you change the limits of integration?

When you introduce u = g(x) in a definite integral, always convert the limits: lower limit becomes g(a) and upper limit becomes g(b). This way you never need to back-substitute. Option (c) is the alternative method and also works, but converting limits is cleaner.

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