Difficulty: Intermediate | Prerequisites: Calculus I (basic integration rules, antiderivatives, chain rule)
Big picture
U-substitution is the single most important integration technique you will learn in Calculus II. It reverses the chain rule of differentiation: wherever a composite function was differentiated, u-substitution lets you undo that composition and reduce the integral to something simpler. Nearly every integration method you meet later (integration by parts, trig substitution, partial fractions) will assume you can already do a u-sub fluently. If you are behind on the chain rule from Calc I, review that first.
TL;DR
U-substitution simplifies integrals by replacing a messy "inner function" with a single variable u, converting the integral into a form you already know how to solve. You identify the inner function, compute du, rewrite everything in terms of u, integrate, then substitute back. For definite integrals, you can convert the limits to u-values and skip the back-substitution entirely.
U-substitution (integration by substitution)
A technique for evaluating integrals by replacing a composite expression with a single variable u, effectively reversing the chain rule. Think of it as: you pick the "inner function" that is making the integral complicated, call it u, and rewrite everything in terms of u until the integral becomes one you recognise.
The substitution u
The new variable you introduce to simplify the integrand. Typically chosen as the "inner function" of a composition, or the expression whose derivative (or a constant multiple of it) already appears elsewhere in the integrand. In simple terms, u is the part you are trying to tame.
du (the differential)
The derivative of u with respect to x, written as du = u'(x) dx. Once you choose u, you compute du and check whether the remaining parts of the integrand match du (up to a constant factor). In simple terms, du tells you what dx "becomes" after you swap in u.
Back-substitution
The final step in an indefinite integral where you replace u with the original expression in x, returning to the original variable. In simple terms, you undo the substitution so the answer is in the same variable as the question.
Change of limits (for definite integrals)
When evaluating a definite integral by substitution, you convert the original x-limits into u-limits using the equation u = g(x). This lets you evaluate entirely in u without ever substituting back. In simple terms, you move the bounds into u-world so you never have to return to x.
Integrand
The function being integrated, i.e. the expression that sits between the integral sign and dx. Think of it as: the thing you are trying to find the antiderivative of.
Even and odd function symmetry (for definite integrals)
If f(x) is odd (f( -x) = -f(x)), then the integral from -a to a equals zero. If f(x) is even (f( -x) = f(x)), then the integral from -a to a equals twice the integral from 0 to a. In simple terms, odd functions cancel themselves out over symmetric intervals; even functions double up.
Step 1: Choose u. Look for a composite function in the integrand. The inner function is usually the right choice for u. Common clues: an expression inside a power, inside a trig function, under a root, or in an exponent.
Step 2: Compute du. Differentiate u with respect to x to get du/dx, then write du = u'(x) dx.
Step 3: Match and rewrite. Check whether the remaining factors in the integrand (everything other than your chosen inner function and dx) match du, possibly up to a constant multiple. If they do, rewrite the entire integral in terms of u and du. If they do not, you may need a different choice of u.
Step 4: Integrate in u. The rewritten integral should be a standard form you recognise (a power rule, an exponential, an inverse trig form, etc.). Evaluate it.
Step 5: Substitute back. Replace u with the original expression in x.
When you set u = g(x), convert the limits: if the original integral runs from x = a to x = b, the new limits are u = g(a) to u = g(b).
Evaluate the antiderivative at the u-limits directly. No back-substitution needed.
If g(b) < g(a), that is fine; just keep the limits in the order they come.
If du differs from what you need by a constant multiple, pull that constant out. For example, if u = x^6, then du = 6x^5 dx, so x^5 dx = (1/6) du. You factor out the 1/6 in front of the integral and proceed.
Sometimes the integrand needs to be split into separate terms before substitution works. For instance, (6 + 3x)/(1 + x^2) splits into 6/(1 + x^2) + 3x/(1 + x^2). The first term is a standard arctangent; the second yields to u = 1 + x^2.
Several forms reduce to inverse trig functions after substitution:
1/(1 + u^2) integrates to arctan(u) + C
1/sqrt(1 - u^2) integrates to arcsin(u) + C
These appear constantly in Calc II problems, often after a substitution strips away the outer layers.
If the integrand is an odd function (f( -x) = -f(x)) and the interval is symmetric about zero (from -a to a), the integral is zero. No computation needed.
If the integrand is an even function (f( -x) = f(x)) over a symmetric interval, the integral equals 2 times the integral from 0 to a.
To check: multiply out the signs. x^2 is even. sin(x) is odd. A product of an even and an odd function is odd. A product of two odd functions is even.
Many trig integrals reduce via substitution:
For powers of sec and tan, try u = sec x (since du = sec x tan x dx). This handles forms like sec^n(x) tan(x) dx.
For expressions involving 1 + cos^2 or 1 + sin^2 in the denominator, try u = cos(x) or u = sin(x) to land on an arctan pattern.
For products like csc(ax) cot(ax), try u = csc(ax) or recognise that the derivative of -csc(ax) is csc(ax) cot(ax) times the chain rule constant.
The substitution formula (indefinite)
\int f(g(x)) \, g'(x) \, dx = \int f(u) \, du \quad \text{where } u = g(x)The substitution formula (definite)
\int_a^b f(g(x)) \, g'(x) \, dx = \int_{g(a)}^{g(b)} f(u) \, duStandard results that appear after substitution
\int \frac{du}{1 + u^2} = \arctan(u) + C\int \frac{du}{\sqrt{1 - u^2}} = \arcsin(u) + C\int e^u \, du = e^u + C\int u^n \, du = \frac{u^{n+1}}{n+1} + C \quad (n \neq -1)\int \frac{du}{u} = \ln|u| + CSymmetry rules for definite integrals
\text{Odd: } \int_{-a}^{a} f(x)\,dx = 0 \quad \text{when } f(-x) = -f(x)\text{Even: } \int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx \quad \text{when } f(-x) = f(x)U-substitution is the mathematical backbone of any problem where a rate of change depends on a composed quantity. Engineers use it to compute total energy stored in a nonlinear spring (where force depends on displacement squared or cubed). Physicists use it to find the total charge flowing through a circuit when current decays exponentially with time, which is exactly the form of the exponential decay integral on the practice sheet (the 115e^( -0.02t) problem). Probability and statistics rely on substitution to evaluate cumulative distribution functions for the normal distribution and other continuous distributions.
Forgetting to convert dx to du. Students often substitute the inner function with u but leave dx unchanged. Every part of the integrand, including the differential, must be rewritten in terms of u and du. If you still have an x anywhere after substituting, something has gone wrong.
Choosing the wrong u. A common instinct is to pick the most complicated-looking piece. That is sometimes right, but the real test is whether du (or a constant multiple of it) appears elsewhere in the integrand. If it does not, your choice of u will not simplify anything.
Forgetting to change the limits in definite integrals. If you switch to u, the limits must also switch to u-values. Evaluating the u-antiderivative at the original x-limits is one of the most common errors on exams.
Dropping the constant factor. When du = 6x^5 dx and you only have x^5 dx, you need a factor of 1/6. Students sometimes set du = x^5 dx directly, losing the 6. Always write out the full du first, then solve for the piece you need.
⚠️ Inverse trig results (arctan, arcsin) after substitution are heavily tested. Know the standard forms cold: 1/(1 + u^2) and 1/sqrt(1 - u^2).
⚠️ Definite integral problems almost always require you to change the limits. Expect at least one on the exam.
⚠️ Problems that combine substitution with trig identities (like sec^n(x) tan(x), or csc(ax) cot(ax)) appear regularly. Practise recognising when u = sec x or u = csc x.
⚠️ Odd/even symmetry can save you significant time on definite integrals over symmetric intervals. If the exam gives you a complicated integrand from -a to a, check symmetry before doing any substitution.
⚠️ Exponential decay integrals (like the 115e^( -0.02t) problem) are a favourite for applied/engineering-flavoured exams. The substitution is mechanical, but the arithmetic with the constants is where marks are lost.
Q: True or false: when using u-substitution on a definite integral, you must always substitute back to x before evaluating.
A: False. If you convert the limits to u-values, you evaluate directly in u and never substitute back.
Q: If u = x^4, what is du?
A: du = 4x^3 dx.
Q: True or false: the integral of 1/(1 + u^2) du is arcsin(u) + C.
A: False. It is arctan(u) + C. The arcsin form is for 1/sqrt(1 - u^2).
Q: If f(x) = x^3 sin(x), is f odd, even, or neither?
A: Even. x^3 is odd, sin(x) is odd, and the product of two odd functions is even.
Q: Fill in the blank: when substituting in a definite integral from x = 0 to x = 1 with u = 1 + 7x, the new limits are u = ___ to u = ___.
A: u = 1 to u = 8.
U-substitution connects directly to the chain rule from Calculus I. Every substitution is the chain rule in reverse: if you can differentiate a composition, you can integrate one by undoing that differentiation. This makes the chain rule the single most important prerequisite.
This topic also sets the stage for integration by parts (which handles products that substitution cannot simplify), trig substitution (a specialised form of u-sub where u is a trig function chosen to eliminate a square root), and partial fractions (which reduces rational functions to simpler pieces, each of which may then need a u-sub).
The even/odd symmetry rules for definite integrals reappear in Fourier series and transform theory, where entire families of integrals vanish by symmetry.
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