Difficulty: Intermediate | Prerequisites: Section 4.4 (net change), antiderivative rules, chain rule for derivatives
U-substitution is the chain rule running backwards. You pick an inner function u = g(x), compute du = g'(x) dx, rewrite the integral entirely in terms of u, and integrate. For definite integrals, you can either convert the bounds to u-values or back-substitute and use the original bounds.
U-substitution (change of variable)
An integration technique where you set u equal to an inner function, replace dx using du = g'(x) dx, and rewrite the integral in a simpler form. Think of it as undoing the chain rule: you are looking for a composition f(g(x)) times g'(x) hiding inside the integrand.
Inner function
The part of a composite expression you choose as u. A good choice is usually the expression inside a power, a root, or a trig function, whose derivative also appears (up to a constant) elsewhere in the integrand.
Bounds conversion (definite integrals)
When using u-sub on a definite integral, you can convert the x-limits to u-limits: if u = g(x), then the lower bound becomes g(a) and the upper bound becomes g(b). This avoids the need to back-substitute.
Odd function property
If f(-t) = -f(t) for all t, then the integral of f(t) from -a to a is zero. The negative area on one side cancels the positive area on the other.
Identify an inner function g(x) and set u = g(x).
Compute du = g'(x) dx and solve for dx (or match du directly to part of the integrand).
Rewrite the entire integral in terms of u and du. No x should remain.
Integrate with respect to u.
Back-substitute u = g(x) to return to x (indefinite integrals) or convert bounds (definite integrals).
3.1: integral of sec^2(x) tan(x) dx
Let u = tan(x), so du = sec^2(x) dx.
Integral becomes: integral of u du = u^2/2 + C = tan^2(x)/2 + C.
3.2: integral of sin(sqrt(x)) / sqrt(x) dx
Let u = sqrt(x), so du = 1/(2 sqrt(x)) dx, which means 2 du = dx/sqrt(x).
Integral becomes: 2 integral of sin(u) du = -2 cos(u) + C = -2 cos(sqrt(x)) + C.
3.3: integral of cos(x) / sqrt(1 + sin(x)) dx
Let u = 1 + sin(x), so du = cos(x) dx.
Integral becomes: integral of u^(-1/2) du = 2 u^(1/2) + C = 2 sqrt(1 + sin(x)) + C.
3.4: integral of 9x sqrt(x - 8) dx
Let u = x - 8, so x = u + 8 and dx = du.
Integral becomes: 9 integral of (u + 8) sqrt(u) du = 9 integral of (u^(3/2) + 8 u^(1/2)) du
= 9 [2u^(5/2)/5 + 16u^(3/2)/3] + C = 18(x - 8)^(5/2)/5 + 48(x - 8)^(3/2) + C.
3.5: integral of x^3 (4 + 5x^4)^5 dx
Let u = 4 + 5x^4, so du = 20x^3 dx, which means x^3 dx = du/20.
Integral becomes: (1/20) integral of u^5 du = u^6/120 + C = (4 + 5x^4)^6 / 120 + C.
4.1: integral from 1 to 2 of x^2 / sqrt(x^3 + 5) dx
Let u = x^3 + 5, so du = 3x^2 dx, meaning x^2 dx = du/3.
Bounds: x = 1 gives u = 6; x = 2 gives u = 13.
= (1/3) integral from 6 to 13 of u^(-1/2) du = (1/3)[2 sqrt(u)] from 6 to 13 = (2/3)(sqrt(13) - sqrt(6)).
4.2: integral from sqrt(pi/2 - 1) to sqrt(pi - 1) of 2x sin(sin(x^2 + 1)) cos(x^2 + 1) dx
First substitution: let w = x^2 + 1, so dw = 2x dx.
Bounds: x = sqrt(pi/2 - 1) gives w = pi/2; x = sqrt(pi - 1) gives w = pi.
Integral becomes: integral from pi/2 to pi of sin(sin(w)) cos(w) dw.
Second substitution: let u = sin(w), so du = cos(w) dw.
Bounds: w = pi/2 gives u = 1; w = pi gives u = 0.
= integral from 1 to 0 of sin(u) du = [-cos(u)] from 1 to 0 = -cos(0) + cos(1) = -1 + cos(1).
4.3: integral from 0 to pi/10 of cos^9(5t) sin(5t) dt
Let u = cos(5t), so du = -5 sin(5t) dt, meaning sin(5t) dt = -du/5.
Bounds: t = 0 gives u = cos(0) = 1; t = pi/10 gives u = cos(pi/2) = 0.
= (-1/5) integral from 1 to 0 of u^9 du = (1/5) integral from 0 to 1 of u^9 du = (1/5)[u^10/10] from 0 to 1 = 1/50.
integral from -7 to 7 of t^3/(11 + 8t^2) dt = 0.
Why: f(t) = t^3/(11 + 8t^2). Check: f(-t) = (-t)^3/(11 + 8t^2) = -t^3/(11 + 8t^2) = -f(t).
The integrand is odd, and the interval [-7, 7] is symmetric about zero. The integral is therefore zero without any computation.
\text{U-Sub (indefinite): } \int f(g(x))\,g'(x)\,dx = \int f(u)\,du \quad\text{where } u = g(x)\text{U-Sub (definite): } \int_a^b f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du\text{Odd function: if } f(-x) = -f(x), \text{ then } \int_{-a}^{a} f(x)\,dx = 0U-substitution is the workhorse technique behind almost every applied integration problem. In physics, computing work done by a variable force often requires a substitution. In probability, converting between distributions uses exactly the same change-of-variable logic.
Students often forget to convert the bounds when doing a definite integral by substitution. If you change the variable to u, you must either convert the limits to u-values or back-substitute before evaluating.
A frequent mistake is leaving x terms in the integral after substituting. Every part of the integrand, including dx, must be expressed in terms of u and du.
Students sometimes choose u poorly. The right u is usually the inner function whose derivative (or a constant multiple of it) appears elsewhere in the integrand.
When the integrand has an extra factor of x that does not match du, students sometimes just drop it. Instead, solve for x in terms of u and substitute (as in Problem 3.4).
U-substitution appears on virtually every Calculus I exam. Expect 2 to 4 integrals requiring it.
Definite integrals with substitution are a favourite exam format because they test two skills at once: the substitution itself and the bounds conversion.
The odd-function shortcut (Problem 5) is a classic exam time-saver. If you spot symmetry, the answer is immediate.
True or false: when using u-sub on a definite integral, you must always back-substitute to x before evaluating. (False. You can convert the bounds to u-values and evaluate directly in u.)
Fill in the blank: if u = sin(x), then du = ______ . (cos(x) dx)
True or false: the integral from -5 to 5 of x^4/(1 + x^2) dx is zero because the integrand is odd. (False. x^4/(1 + x^2) is even, not odd, so the integral is not zero.)
Fill in the blank: the key to choosing a good u is finding an inner function whose ______ appears elsewhere in the integrand. (derivative)
True or false: after substitution, the integrand should contain only u and du, with no x remaining. (True.)
Q: Evaluate the integral of 3x^2 cos(x^3) dx.
A: Let u = x^3, du = 3x^2 dx. Integral becomes integral of cos(u) du = sin(u) + C = sin(x^3) + C.
Q: Evaluate the integral from 0 to 1 of x/(1 + x^2)^2 dx.
A: Let u = 1 + x^2, du = 2x dx. Bounds: u(0) = 1, u(1) = 2. Integral = (1/2) integral from 1 to 2 of u^(-2) du = (1/2)[-1/u] from 1 to 2 = (1/2)(-1/2 + 1) = 1/4.
Q: Is the integral from -3 to 3 of x^5 sin(x)/(1 + x^4) dx equal to zero? Why or why not?
A: Check: f(-x) = (-x)^5 sin(-x)/(1 + x^4) = -x^5 (-sin(x))/(1 + x^4) = x^5 sin(x)/(1 + x^4) = f(x). The function is even, not odd, so the integral is not zero.
Q: Evaluate the integral of e^(2x) / (1 + e^(2x)) dx.
A: Let u = 1 + e^(2x), du = 2e^(2x) dx. Integral = (1/2) integral of du/u = (1/2) ln|u| + C = (1/2) ln(1 + e^(2x)) + C.
U-substitution is the inverse of the chain rule from Section 3.6. Mastering it is essential for integration by parts (Calc II), which handles cases where u-sub alone is not enough. The odd/even function symmetry properties also come back in Fourier series and other later courses. Section 5.1 (area between curves) will frequently require u-sub to evaluate the resulting integrals.
u-substitution, change of variable, integration by substitution, reverse chain rule, definite integral substitution, bounds conversion, odd function integral, even function integral, MTH 132, Calculus I, Michigan State University, Section 4.5, recitation 14, trig substitution, nested substitution, double substitution