The chain-rule-in-reverse recognition from earlier in this chapter turns into an actual computational technique here. An integrand can be spotted as an outer function applied to an inner one, times the inner function's derivative. Once that shape is spotted, renaming the inner function u turns the whole integral into something the basic antiderivative rules can handle directly.
It's a disguise, not a shortcut. The integral \int 2x\cos(x^2)\,dx looks nothing like a basic antiderivative. But underneath the disguise, it's just \int \cos(u)\,du wearing an x^2-shaped costume. Renaming u=x^2 doesn't change the math at all. It just removes the costume so the simple rule underneath becomes visible again. On a definite integral, the two boundary numbers were measured in the ORIGINAL costume, in terms of x. They have to be re-measured in terms of u too, or the finish line doesn't match the start.
U-substitution is the name for turning the chain-rule-in-reverse recognition from earlier in this chapter into an actual computational technique: renaming an integrand's inner function u so the basic antiderivative rules can handle what's left directly. The mechanical steps: pick u to be the inner function, compute du = g'(x)\,dx, and rewrite the ENTIRE integral, every x, in terms of u alone. If any x is left over after that rewrite, either u was the wrong choice or the integral isn't a chain-rule shape at all. A clean substitution leaves nothing but u's and du's. On a definite integral, the bounds must be converted too. Evaluate u=g(x) at both original x-bounds to get the matching u-bounds. That lets the whole problem finish in terms of u without ever converting back to x at the end.
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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Which u actually works?
Try each candidate — only one leaves nothing extra behind once du is accounted for.
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Worked example
u-substitution on a definite integral
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1
Evaluate \displaystyle\int_0^2 x\sqrt{x^2+1}\,dx.
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2the inner function under the root is the natural choice for u
Let u = x^2+1, so du = 2x\,dx, i.e. x\,dx = \frac{1}{2}du.
the inner function under the root is the natural choice for u
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3
Convert the bounds: at x=0, u=1; at x=2, u=5. The integral becomes \int_1^5 \sqrt{u}\cdot\frac{1}{2}\,du = \frac{1}{2}\int_1^5 u^{1/2}\,du.
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4
Antidifferentiate and evaluate in u directly. There's no need to convert back to x: \frac{1}{2}\cdot\frac{2}{3}u^{3/2}\Big|_1^5 = \frac{1}{3}\left(5^{3/2}-1\right). \boxed{\int_0^2 x\sqrt{x^2+1}\,dx = \frac{1}{3}\left(5\sqrt5-1\right)}
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Where this goes next
Every technique chapter ahead
Integration by parts, trig substitution, and partial fractions all still lean on u-substitution as a final cleanup step. None of them replace it. They all assume it's already automatic.
Check your understanding (counts toward your progress)
Evaluate \displaystyle\int_0^6 x\sqrt{x^2+7}\,dx, rounded to 3 decimal places.
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
With u=x^2+7 (bounds 7\to43), the integral is \dfrac{1}{3}\left(43^{3/2}-7^{3/2}\right) \approx 87.817.
Correct answer
Also accepted:
Why
Let u=x^2+7; rewrite both the integrand and the bounds in terms of u.
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.
Find \displaystyle\int 2x\cos(x^2)\,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=x^2, du=2x\,dx — the integral becomes \int\cos u\,du = \sin u = \sin(x^2), plus C.
Correct answer
Also accepted:
Why
Let u=x^2.
Find \displaystyle\int 12x\sin(6x^2)\,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=6x^2, du=12x\,dx — the integral becomes \int\sin u\,du = -\cos u = -\cos(6x^2), plus C.
Correct answer
Also accepted:
Why