Source: Chapter 5, Section 5.5 | Calculus (Texas A&M University)
Tags: u-substitution, integration by substitution, change of variable, chain rule in reverse, definite integral substitution, indefinite integral substitution, du, inner function
U-substitution is the chain rule run in reverse. When an integrand has the form f(g(x)) · g′(x), you let u = g(x), replace g′(x) dx with du, and integrate f(u) du instead. For definite integrals, you can either change the limits to match u, or compute the indefinite integral first and plug in the original x-limits at the end.
U-substitution (integration by substitution)
A technique for evaluating integrals where you identify an "inner function" u = g(x), compute du = g′(x) dx, and rewrite the integral entirely in terms of u. The reverse of the chain rule for differentiation.
du
The differential of u. If u = g(x), then du = g′(x) dx. In substitution, you replace the g′(x) dx portion of the integrand with du.
Inner function
The function you choose as u. Typically, it is the part of the integrand that is "inside" another function, or the part whose derivative also appears in the integrand.
The chain rule says: d/dx [F(g(x))] = F′(g(x)) · g′(x) = f(g(x)) · g′(x).
Reading this backwards: ∫ f(g(x)) g′(x) dx = F(g(x)) + C = F(u) + C = ∫ f(u) du.
Substitution is just the chain rule, unwound.
Indefinite integrals:
∫ f(g(x)) g′(x) dx = ∫ f(u) du, where u = g(x), du = g′(x) dx.
Definite integrals, Method I (change 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 go back to x.
Definite integrals, Method II (back-substitute):
Compute the indefinite integral ∫ f(g(x)) g′(x) dx first (in terms of x), then evaluate at the original limits a and b.
Both methods give the same answer. Pick whichever feels cleaner for the problem at hand.
The general strategy:
Look for a composite function in the integrand, something of the form f(g(x)).
Let u = g(x), the "inner" part.
Compute du = g′(x) dx.
Check that g′(x) dx (or a constant multiple of it) appears in the integrand.
Rewrite everything in terms of u and du. If any x remains, the substitution is incomplete.
Sometimes guessing the right u takes practice. If your first choice doesn't work cleanly, try another.
∫ e^{3x+2} dx (Example 5.5.1)
Let u = 3x + 2, so du = 3 dx, meaning dx = du/3.
∫ eᵘ (du/3) = (1/3) eᵘ + C = (1/3) e^{3x+2} + C.
∫ sin(x) cos(x) dx (Example 5.5.3)
Let u = sin x, so du = cos x dx.
∫ u du = u²/2 + C = sin²(x)/2 + C.
(You could also let u = cos x, which gives −cos²(x)/2 + C. These differ by a constant, so both are correct.)
∫₀¹ e^{3x+2} dx (Example 5.5.6)
Method I: u = 3x+2, du = 3dx. When x = 0, u = 2. When x = 1, u = 5.
∫₂⁵ eᵘ (du/3) = (1/3)[eᵘ]₂⁵ = (1/3)(e⁵ − e²).
Method II: anti-derivative is (1/3)e^{3x+2}. Evaluate: (1/3)e⁵ − (1/3)e² = (1/3)(e⁵ − e²). Same answer.
∫₀¹ 1/(x+1) dx (Example 5.5.7)
Let u = x + 1, du = dx. When x = 0, u = 1. When x = 1, u = 2.
∫₁² (1/u) du = [ln|u|]₁² = ln 2 − ln 1 = ln 2.
∫₁ᵉ (ln x)/x dx (Example 5.5.8)
Let u = ln x, so du = (1/x) dx. When x = 1, u = 0. When x = e, u = 1.
∫₀¹ u du = [u²/2]₀¹ = 1/2.
∫₀¹ (2x − 1)¹⁰⁰ dx (Example 5.5.10)
Let u = 2x − 1, du = 2 dx, so dx = du/2. When x = 0, u = −1. When x = 1, u = 1.
(1/2) ∫₋₁¹ u¹⁰⁰ du.
Notice: u¹⁰⁰ is an even function, and the interval is symmetric. So this equals (1/2) · 2 ∫₀¹ u¹⁰⁰ du = ∫₀¹ u¹⁰⁰ du = [u¹⁰¹/101]₀¹ = 1/101.
(This is a nice example of combining substitution with even/odd shortcuts.)
∫ tan x dx (Example 5.5.11)
Rewrite: ∫ sin x / cos x dx. Let u = cos x, du = −sin x dx.
∫ (−1/u) du = −ln|u| + C = −ln|cos x| + C.
This can also be written as ln|sec x| + C.
∫ cos(√x) / √x dx (Example 5.5.12)
Let u = √x = x^{1/2}, so du = (1/2)x^{−1/2} dx = 1/(2√x) dx.
Then (1/√x) dx = 2 du.
∫ cos(u) · 2 du = 2 sin u + C = 2 sin(√x) + C.
∫₀⁴ x/√(1+2x) dx (Example 5.5.13)
Let u = 1 + 2x, so du = 2 dx and x = (u − 1)/2. When x = 0, u = 1. When x = 4, u = 9.
∫₁⁹ [(u−1)/2] / √u · (du/2) = (1/4) ∫₁⁹ (u−1)/√u du = (1/4) ∫₁⁹ (u^{1/2} − u^{−1/2}) du.
Integrate: (1/4) [(2/3)u^{3/2} − 2u^{1/2}]₁⁹ = (1/4){[(2/3)(27) − 2(3)] − [(2/3)(1) − 2(1)]}
= (1/4){[18 − 6] − [2/3 − 2]} = (1/4){12 − (−4/3)} = (1/4)(40/3) = 10/3.
∫ (1+x)/(1+x²) dx (Example 5.5.14)
Split the fraction: 1/(1+x²) + x/(1+x²).
The first part: ∫ 1/(1+x²) dx = arctan(x) + C.
The second part: let u = 1 + x², du = 2x dx. So ∫ x/(1+x²) dx = (1/2) ln|1+x²| + C.
Combined: arctan(x) + (1/2) ln(1+x²) + C. (The absolute value is not needed since 1+x² is always positive.)
Substitution rule:
∫ f(g(x)) g′(x) dx = ∫ f(u) du, with u = g(x)
Definite integral version:
∫ₐᵇ f(g(x)) g′(x) dx = ∫_{g(a)}^{g(b)} f(u) du
Key result from substitution:
∫ tan x dx = −ln|cos x| + C = ln|sec x| + C
Logarithm review (Review 5.5.9):
ln eᵃ = a
ln(ab) = ln a + ln b
ln aᵇ = b ln a
ln e = 1, ln 1 = 0
ln a is defined only for a > 0
⚠️ After substituting, no x should remain in the integral. If it does, either solve for x in terms of u (as in the x/√(1+2x) example) or reconsider your choice of u.
⚠️ When using Method I for definite integrals (changing limits), you must convert the limits to u-values. A very common error is keeping the original x-limits with a u-integrand.
⚠️ Don't forget the factor from du. If du = 3 dx, then dx = du/3, and you need that 1/3 out front.
⚠️ ∫ tan x dx = ln|sec x| + C is a result worth memorising. It comes up frequently and the derivation via substitution is a common exam question.
⚠️ Splitting fractions (as in (1+x)/(1+x²)) is a useful technique that pairs well with substitution. If the integrand doesn't fit a single substitution, try algebraic manipulation first.
Q: Evaluate ∫ 2x · e^{x²} dx.
A: Let u = x², du = 2x dx. The integral becomes ∫ eᵘ du = eᵘ + C = e^{x²} + C.
Q: Evaluate ∫₀^{π/2} cos x · e^{sin x} dx.
A: Let u = sin x, du = cos x dx. When x = 0, u = 0. When x = π/2, u = 1. Integral: ∫₀¹ eᵘ du = e − 1.
Q: What goes wrong if you try u = x² for ∫ x³ · e^{x²} dx?
A: du = 2x dx, so you can replace 2x dx with du. But x³ dx = x² · x dx = u · (du/2). The integral becomes (1/2) ∫ u eᵘ du, which requires integration by parts (a different technique). The substitution alone doesn't finish the problem.
Q: Evaluate ∫ sec²(3x) dx.
A: Let u = 3x, du = 3 dx. Integral: (1/3) ∫ sec² u du = (1/3) tan u + C = (1/3) tan(3x) + C.
Q: Find ∫₀¹ (2x−1)¹⁰⁰ dx using substitution.
A: Let u = 2x − 1, du = 2 dx. Limits: u(0) = −1, u(1) = 1. Integral: (1/2) ∫₋₁¹ u¹⁰⁰ du. Since u¹⁰⁰ is even: (1/2)(2) ∫₀¹ u¹⁰⁰ du = [u¹⁰¹/101]₀¹ = 1/101.
u-substitution, integration by substitution, change of variable, reverse chain rule, du, inner function, composite function integration, definite integral substitution, changing limits of integration, back-substitution, tan x integral, logarithm rules, splitting fractions, algebraic manipulation before integration