Double and triple integrals don't introduce a new way to antidifferentiate. Every inner integral that comes up along the way still has to be solved with the same techniques already in hand. Of those, two carry almost all of the weight ahead: u-substitution and integration by parts. U-substitution handles an integrand that's secretly a chain rule run backward. Integration by parts handles one that's secretly a product rule run backward.
u-substitution undoes a chain rule. Find the "inner function" whose derivative is lurking in the integral, and rename it u. Once you do, the outer function suddenly looks ordinary. Integration by parts undoes a product rule instead. Split the integrand into two factors. Differentiate the one that gets simpler with each derivative, like x, which becomes 1. Integrate the other factor, the one that doesn't mind being integrated repeatedly, like \cos x or e^x.
This course's own inner integrals lean on two integration techniques almost exclusively: u-substitution, which undoes a chain rule, and integration by parts, which undoes a product rule. The hard part of u-substitution was never computing du. It's recognizing which piece of the integrand to call u in the first place: the piece whose derivative is, up to a constant, sitting elsewhere in the integral.
Integration by parts asks a related but distinct question: which factor to call u (differentiate it) and which to call dv (integrate it). LIATE gives a rough order of preference for u: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. Each later category in that list tends to integrate more easily than it differentiates.
Formula
u-Substitution
also known as: Change of Variables, Reverse Chain Rule
Recalled from Calculus I — Differential & Integral Calculus · U-substitution: reversing the chain rule — first taught there.
If u = g(x) with g differentiable and f continuous on the range of g, then \int f(g(x))\,g'(x)\,dx = \int f(u)\,du. For a definite integral, the bounds transform along with the variable: \int_a^b f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du.
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Formula
Integration by Parts
also known as: the product rule run in reverse
Recalled from Calculus II — Integral Calculus & Infinite Series · Integration by parts — first taught there.
For differentiable u=u(x) and v=v(x): \int u\,dv = uv - \int v\,du.
Start from the product rule, \dfrac{d}{dx}[uv] = u\,\dfrac{dv}{dx} + v\,\dfrac{du}{dx}, and integrate both sides with respect to x: uv = \int u\,\frac{dv}{dx}\,dx + \int v\,\frac{du}{dx}\,dx = \int u\,dv + \int v\,du. Solving for \int u\,dv gives the formula directly. No information is added or lost — it's the product rule read from right to left, with the two terms it produces separated onto opposite sides of the equation.
Which piece is u?
Click each candidate — only one leaves nothing but a clean integral in u behind.
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Every antiderivative fits the same slope field
Drag C — every curve in the family stays tangent to the same field, the family integration by parts produces below.
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Worked example
Integration by parts
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1
Evaluate \displaystyle\int x\cos x\,dx.
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2the opposite choice, u = cos x, would make du and v both messier, not simpler
By LIATE, let u=x (algebraic, gets simpler) and dv=\cos x\,dx (trigonometric, easy to integrate). Then du=dx and v=\sin x.
the opposite choice, u = cos x, would make du and v both messier, not simpler
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3
Apply the formula \int u\,dv = uv - \int v\,du: \int x\cos x\,dx = x\sin x - \int \sin x\,dx
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4
\boxed{\int x\cos x\,dx = x\sin x + \cos x + C}
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Where this goes next
Chapter 7: Double integrals
An inner integral inside a double integral is an ordinary single-variable integral, with the other variable frozen as a constant. The very first double integral this course computes will need exactly one of these two techniques. It applies to one variable at a time.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus II — Integral Calculus & Infinite Series
Find \displaystyle\int \dfrac{x}{x^2+4}\,dx.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
\int \tfrac12\cdot\tfrac{1}{u}\,du = \tfrac12\ln|u| = \tfrac12\ln(x^2+4), plus C.
Correct answer
Also accepted:
Why
Let u=x^2+4, so du=2x\,dx.
From Calculus II — Integral Calculus & Infinite Series
Find \displaystyle\int \sin^3 x\cos x\,dx.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
With u=\sin x, du=\cos x\,dx — \int u^3\,du = \tfrac{u^4}{4} = \tfrac{\sin^4 x}{4}, plus C.
Correct answer
Also accepted:
Why
Let u=\sin x.
From Calculus II — Integral Calculus & Infinite Series
Find \displaystyle\int (2x+1)^5\,dx.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
\int u^5 \cdot \tfrac12\,du = \tfrac{1}{12}u^6 = \tfrac{(2x+1)^6}{12}, plus C.
Correct answer
Also accepted:
Why
Let u=2x+1, so du=2\,dx.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus II — Integral Calculus & Infinite Series
Find \displaystyle\int x\sin(2x)\,dx.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
-\tfrac x2\cos2x - \int\left(-\tfrac12\cos2x\right)dx = -\tfrac x2\cos2x+\tfrac14\sin2x, plus C.
Correct answer
Also accepted:
Why
Let u=x, dv=\sin(2x)\,dx, so v=-\tfrac12\cos(2x).
From Calculus II — Integral Calculus & Infinite Series
Evaluate \displaystyle\int_0^{\pi/2} x\sin x\,dx.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
[-x\cos x+\sin x]_0^{\pi/2} = (0+1)-(0+0) = 1.
Correct answer
Also accepted:
Why
Let u=x, dv=\sin x\,dx, then evaluate [-x\cos x+\sin x] at the bounds.
From Calculus II — Integral Calculus & Infinite Series
Find \displaystyle\int x e^{2x}\,dx.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
u=x, dv=e^{2x}dx gives \tfrac{x}{2}e^{2x}-\int\tfrac12e^{2x}dx = \tfrac x2 e^{2x}-\tfrac14e^{2x}, plus C.
Correct answer
Also accepted:
Why
Let u=x, dv=e^{2x}\,dx, so v=\tfrac12 e^{2x}.