Difficulty: Introductory to Intermediate | Prerequisites: Calculus I (basic differentiation rules, antiderivatives, power rule for integration).
U-substitution is the single most important technique you learn in Calculus II for evaluating integrals. It reverses the chain rule of differentiation: wherever you spot a composite function and its inner derivative lurking in the integrand, you can simplify the whole expression into something elementary. Nearly every integration method you meet later (integration by parts, trig substitution, partial fractions) assumes you can already do a u-sub fluently, so this is the foundation everything else sits on. If you can differentiate confidently using the chain rule, you have the prerequisite; if that still feels shaky, revisit it before going further.
Pick a piece of the integrand to call u, rewrite everything (including dx) in terms of u and du, integrate the simpler expression, then substitute back. The trick is choosing the right u, which is nearly always the "inner function" of a composition whose derivative also appears in the integrand.
U-substitution (integration by substitution)
A technique for evaluating integrals by introducing a new variable u = g(x) so that a composite integrand simplifies to a standard form. Think of it as running the chain rule in reverse.
Integrand
The expression being integrated, i.e. everything between the integral sign and the dx. In simple terms, it is the function you are finding the area under.
Differential (du)
The derivative of u with respect to x, multiplied by dx. Formally, if u = g(x) then du = g'(x) dx. In simple terms, du is what dx "becomes" after the substitution.
Inner function
In a composite expression like f(g(x)), the inner function is g(x). This is almost always what you set u equal to.
Chain rule (reverse)
The differentiation rule stating that d/dx [f(g(x))] = f'(g(x)) · g'(x). U-substitution exploits this in reverse: if you see f'(g(x)) · g'(x) inside an integral, substituting u = g(x) collapses it to ∫ f'(u) du.
Constant of integration (C)
The arbitrary constant added to every indefinite integral, representing the family of all antiderivatives. Forgetting it is the most common mark lost on exams.
Identify the inner function. Look for a composite expression in the integrand. The argument of the outermost operation (the thing inside the trig function, inside the exponent, under the root, in the denominator) is usually your u.
Set u and find du. Write u = g(x), differentiate to get du = g'(x) dx, then solve for dx (or for the leftover x-expression times dx).
Rewrite the integral entirely in terms of u. Every x must vanish. If it does not, your choice of u may be wrong.
Integrate in terms of u. Apply standard antiderivative rules.
Substitute back. Replace u with the original expression in x. Add C.
Example 1: Trig function with a linear inner argument
\int \cos(6x)\,dxLet u = 6x, so du = 6 dx, meaning dx = (1/6) du.
The integral becomes (1/6) ∫ cos(u) du = (1/6) sin(u) + C.
Substitute back: (1/6) sin(6x) + C.
Pattern: when the inner function is linear (ax + b), you simply divide by the coefficient a.
Example 2: Power function under a root
\int x^2 \sqrt{x^3 + 1}\,dxLet u = x³ + 1, so du = 3x² dx, meaning x² dx = (1/3) du.
The integral becomes (1/3) ∫ √u du = (1/3) · (2u^(3/2))/3 + C = (2/9)(x³ + 1)^(3/2) + C.
Result: (2/9)(x³ + 1)^(3/2) + C.
Pattern: the x² sitting outside is exactly (up to a constant) the derivative of the expression under the root. That is your cue.
Example 3: Rational function producing a logarithm
\int \frac{x^3}{x^4 - 8}\,dxLet u = x⁴ - 8, so du = 4x³ dx, meaning x³ dx = (1/4) du.
The integral becomes (1/4) ∫ (1/u) du = (1/4) ln|u| + C.
Result: (1/4) ln|x⁴ - 8| + C.
Pattern: when the numerator is (a constant multiple of) the derivative of the denominator, the integral is a logarithm.
Example 4: A binomial raised to a power
\int (4 - 4x)^6\,dxLet u = 4 - 4x, so du = -4 dx, meaning dx = -(1/4) du.
The integral becomes -(1/4) ∫ u⁶ du = -(1/4) · u⁷/7 + C = -(4 - 4x)⁷/28 + C.
Result: -(4 - 4x)⁷ / 28 + C.
Pattern: same linear-inner-function shortcut as Example 1, but with a power instead of a trig function.
Example 5: A power of ln(x)
\int \frac{(\ln x)^{12}}{x}\,dxLet u = ln x, so du = (1/x) dx. The (1/x) dx in the integrand is exactly du.
The integral becomes ∫ u¹² du = u¹³/13 + C.
Result: (ln x)¹³ / 13 + C.
Pattern: any time you see 1/x paired with a function of ln x, set u = ln x.
Example 6: Fractional-power trig substitution
\int \sqrt[26]{\cot x}\;\csc^2 x\,dxLet u = cot x, so du = -csc²x dx, meaning csc²x dx = -du.
The integral becomes -∫ u^(1/26) du = -u^(27/26) / (27/26) + C = -(26/27)(cot x)^(27/26) + C.
Result: -(26/27)(cot x)^(27/26) + C.
Pattern: csc²x is the derivative of cot x (up to sign). Whenever a trig function and its derivative appear together, that is a u-sub.
Example 7: Double-angle and a shifted denominator
\int \frac{\sin(2x)}{23 + \cos^2 x}\,dxLet u = 23 + cos²x. Then du = -2 cos x sin x dx = -sin(2x) dx (using the identity sin(2x) = 2 sin x cos x).
So sin(2x) dx = -du, and the integral becomes -∫ (1/u) du = -ln|u| + C.
Result: -ln|23 + cos²x| + C.
(The absolute value bars are technically redundant here because 23 + cos²x > 0 always, but writing them is good practice.)
Pattern: recognising that sin(2x) is the derivative of -cos²x (via the double-angle identity) is the key insight.
The substitution rule (general form)
\int f(g(x))\,g'(x)\,dx = \int f(u)\,du \quad\text{where } u = g(x)Common derivative pairs to spot
Inner function u | Its derivative du/dx | What to look for in the integrand |
|---|---|---|
ax + b | a | A constant coefficient on dx |
xⁿ (e.g. x³ + 1) | nxⁿ⁻¹ | An x-power one degree lower |
sin x | cos x | cos x alongside a function of sin x |
cos x | -sin x | sin x alongside a function of cos x |
tan x | sec²x | sec²x alongside a function of tan x |
cot x | -csc²x | csc²x alongside a function of cot x |
eˣ | eˣ | eˣ alongside itself |
ln x | 1/x | 1/x alongside a function of ln x |
Useful identity for these problems
\sin(2x) = 2\sin x \cos xThis identity is what makes Example 7 work: it lets you recognise sin(2x) as the derivative of -cos²x.
U-substitution is how engineers and physicists evaluate the integrals that describe energy stored in a spring (where force depends on displacement), charge accumulation in a capacitor (where current depends on voltage through a nonlinear function), and heat transfer through materials with temperature-dependent conductivity. Any time a rate of change is a composite function of the variable you are integrating over, u-substitution is the tool that makes the maths tractable.
Students often forget to convert dx into du. Setting u and rewriting the function is not enough; the differential must change too, or the integral is wrong.
Students sometimes leave x-terms in the integral after substituting. If any x remains and cannot be expressed in terms of u, the substitution is incomplete or the wrong u was chosen.
Students frequently drop the constant multiplier that comes from solving du = g'(x) dx for dx. For instance, if du = 3x² dx, then x² dx = (1/3) du, and that 1/3 must appear in front of the integral.
Students occasionally forget to substitute back from u to x at the end. The final answer to an indefinite integral in x must be written in terms of x, not u.
⚠️ U-substitution appears on virtually every Calculus II exam, often multiple times. Expect at least two or three problems.
⚠️ Exams love to test whether you can spot the substitution when the derivative of the inner function is present but scaled by a constant (as in Examples 1 through 4).
⚠️ The logarithmic result (Example 3) is a favourite: if the numerator is the derivative of the denominator, the answer is ln|denominator| + C.
⚠️ Forgetting the constant of integration C on an indefinite integral typically costs a mark every time.
True or false: in u-substitution, you must rewrite every part of the integrand (including dx) in terms of u and du. True.
Fill in the blank: if u = x⁴ - 8, then du = ____. 4x³ dx.
True or false: the integral ∫ f'(g(x)) dx can always be evaluated by setting u = g(x). False. You also need g'(x) to appear in the integrand (or be a constant).
Fill in the blank: ∫ cos(6x) dx = ____. (1/6) sin(6x) + C.
True or false: when the numerator of a fraction is the derivative of the denominator, the integral is a natural logarithm. True.
Q: Evaluate ∫ sin(5x) dx.
A: Let u = 5x, du = 5 dx. The integral becomes (1/5) ∫ sin(u) du = -(1/5) cos(5x) + C.
Q: Evaluate ∫ x · e^(x²) dx.
A: Let u = x², du = 2x dx. The integral becomes (1/2) ∫ eᵘ du = (1/2) e^(x²) + C.
Q: Evaluate ∫ (2x)/(x² + 9) dx.
A: Let u = x² + 9, du = 2x dx. The integral becomes ∫ (1/u) du = ln|x² + 9| + C. (The absolute value is technically redundant since x² + 9 > 0 always.)
Q: What is the first thing you should look for when deciding whether u-substitution applies?
A: A composite function whose inner function's derivative (up to a constant factor) also appears in the integrand.
Q: A student writes ∫ (3x + 1)⁴ dx = (3x + 1)⁵/5 + C. What did they do wrong?
A: They forgot to account for the derivative of the inner function. Since d/dx(3x + 1) = 3, the correct answer is (3x + 1)⁵ / 15 + C.
U-substitution connects directly to the chain rule from Calculus I; mastering one deepens your understanding of the other. It is also the first step toward integration by parts (Calculus II), where the integral is split differently but the substitution logic is similar. Later, when you meet differential equations, separable equations are solved by exactly this kind of rearrangement and integration.
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