u-substitution
The chain rule differentiates any composition. Can it be run backward, so that an integrand built from a function of a function and the inner derivative can be antidifferentiated in one step?
Where you are 16 in Integrals
When you reach for it
The rules in Antiderivatives and indefinite integrals and Antiderivatives of trig, exponential and log functions antidifferentiate x^n, \sin x, e^x and \displaystyle \frac{1}{x} when the input is x itself, or at most kx. They say nothing about \displaystyle 2x\sqrt{x^2 + 9} or x^2\cos(x^3 + 1). Yet these integrands have a recognizable look: a function of an inside expression, multiplied by the derivative of that inside expression. That is exactly what the chain rule produces when it differentiates a composition. Substitution reads the chain rule backward.
You reach for it when an integrand contains a composition whose inside, or something close to it, has its derivative sitting in the integrand as a factor, up to a constant. With the second fundamental theorem it also evaluates definite integrals, and it is the most used technique of integration after the basic table.
The procedure
Day 1: the chain rule read backward
Suppose F is an antiderivative of f, so F' = f. By the chain rule,
Read as a statement about antiderivatives, this is the whole method.
Theorem (substitution rule). Let g be differentiable on an interval I with g' continuous there, and let f be continuous on an interval J that contains every value g(x) for x in I. If F is an antiderivative of f on J, then on I
The proof is the line above it: the derivative of the right side is the integrand.
The differential as bookkeeping. Give the inside a name, u = g(x). Its differential, as in Linear approximation and differentials, is du = g'(x)\,dx. In the integrand f\big(g(x)\big)\,g'(x)\,dx, the part g(x) becomes u and the part g'(x)\,dx becomes du:
Putting u = g(x) back gives F\big(g(x)\big) + C, the theorem’s answer. The notation does the chain rule for us: replacing g'(x)\,dx by du is legitimate because the theorem says the result is correct, and differentiating the final answer confirms it every time.
Choosing u. Look for a composition, an expression sitting inside a power, a root, a trig function, an exponential or a denominator, whose derivative also appears as a factor. Then every x in the integrand must be accounted for: what is left after u is named must be exactly du, or a constant multiple of it. If any x is left over that cannot be written in terms of u, the choice has failed for that integrand.
Constant factors. If u = x^3 + 1, then du = 3x^2\,dx, and an integrand that holds only x^2\,dx needs \displaystyle x^2\,dx = \frac{1}{3}\,du. This is allowed because a constant moves through an integral sign, \int k\,h(x)\,dx = k\int h(x)\,dx, the constant multiple rule from Antiderivatives and indefinite integrals. A variable factor does not move through it (Example 2).
Checking. An antiderivative is checked by differentiating it. After a substitution this is the chain rule again, and the inside derivative it produces must cancel whatever constant was adjusted.
Day 2: the shapes substitution handles
The same theorem, with different outer functions f, gives a short list of shapes worth recognizing.
| Integrand | Choice | Antiderivative |
|---|---|---|
| \big(g(x)\big)^n g'(x), n \ne -1 | u = g(x) | \dfrac{\big(g(x)\big)^{n+1}}{n+1} + C |
| \dfrac{g'(x)}{g(x)} | u = g(x) | \ln\lvert g(x) \rvert + C |
| e^{g(x)}g'(x) | u = g(x) | e^{g(x)} + C |
| \sin^n x\cos x | u = \sin x | \dfrac{\sin^{n+1} x}{n+1} + C |
| \dfrac{x}{x^2 + c} | u = x^2 + c | \dfrac{1}{2}\ln\lvert x^2 + c \rvert + C |
The logarithm row holds on any interval where g(x) \ne 0, since \ln\lvert u \rvert is an antiderivative of \displaystyle \frac{1}{u} only away from u = 0. The bars can be dropped when g(x) > 0 throughout, as for x^2 + 5.
Earlier lessons checked some answers by differentiating and pointed here: \displaystyle \int e^{kx}\,dx = \frac{1}{k}e^{kx} + C and \displaystyle \int \frac{dx}{ax + b} = \frac{1}{a}\ln\lvert ax + b \rvert + C (Antiderivatives of trig, exponential and log functions, Dividing before integrating) are the substitutions u = kx and u = ax + b, and the shifted arctangent of Completing the square inside an integral is the substitution \displaystyle u = \frac{x - h}{k}.
An extra x. Sometimes u and du account for everything except a stray factor of x. If u = x + 3, then x = u - 3, and the stray x can be rewritten in terms of u. The integrand then holds only u, and the substitution goes through (Example 5). The test is always the same: after the change, no x may remain.
When nothing works. Substitution needs the derivative of the inside to be present. For \cos(x^2) there is no factor x to pair with u = x^2, and rewriting dx as \displaystyle \frac{du}{2\sqrt{u}} only moves the difficulty into the new integrand. Other methods, such as integration by parts and partial fractions, handle other shapes. Some continuous functions, \cos(x^2) and e^{-x^2} among them, have no antiderivative built from the usual functions at all; this is a theorem, stated here without proof. Their definite integrals still exist and are estimated numerically (The trapezoidal rule and Simpson's rule).
Day 3: substitution in a definite integral
Theorem (substitution in a definite integral). Let g' be continuous on [a, b], and let f be continuous on an interval containing every value of g on [a, b]. Then
Proof. Let F be an antiderivative of f. By the chain rule, F\big(g(x)\big) is an antiderivative of the left integrand on [a, b], so by the second fundamental theorem the left side is F\big(g(b)\big) - F\big(g(a)\big). The same theorem applied to the right side gives F(u)\Big|_{g(a)}^{g(b)}, the same number. \blacksquare
This gives two routes to a definite integral, and they agree:
- Change the limits. Replace each x-limit by its u-value, g(a) and g(b), and evaluate in u. There is no need to change back to x.
- Back-substitute. Find the antiderivative in x first, as on Day 1, and evaluate it at the original limits a and b.
The two must not be mixed: an integral written in u with limits that are still x-values gives a wrong number (Exercise 24). If g(a) > g(b), the u-integral runs from the larger number to the smaller, which is allowed; by the convention of Signed area and integrals from geometry it equals the negative of the integral with the limits swapped.
Symmetric intervals. Properties of definite integrals showed with midpoint sums that for f continuous on [-a, a], \int_{-a}^{a} f(x)\,dx = 0 when f is odd and 2\int_0^a f(x)\,dx when f is even. Substitution gives a second proof. Split at 0, and in the left piece put u = -x, du = -dx, so that x = -a becomes u = a and x = 0 becomes u = 0:
For odd f this is -\int_0^a f(u)\,du, which cancels the right piece \int_0^a f(x)\,dx. For even f it equals the right piece, and the total is twice it. The letter used for the variable does not change the value of a definite integral, which is why \int_0^a f(u)\,du and \int_0^a f(x)\,dx are the same number.
Worked examples
Day 1: the derivative of the inside is already there
Find (a) \displaystyle \int 2x\sqrt{x^2 + 9}\,dx and (b) \int (4 - 3x)^5\,dx. Check each by differentiating.
Solution(a) The root holds x^2 + 9, and its derivative 2x is the other factor. Let u = x^2 + 9, so du = 2x\,dx. Then \displaystyle \sqrt{x^2 + 9} becomes u^{1/2} and 2x\,dx becomes du, with no x left:
Check: by the chain rule, \displaystyle \frac{2}{3} \cdot \frac{3}{2}\left(x^2 + 9\right)^{1/2} \cdot 2x = 2x\sqrt{x^2 + 9}.
(b) Let u = 4 - 3x, so du = -3\,dx and \displaystyle dx = -\frac{1}{3}\,du. The constant \displaystyle -\frac{1}{3} moves outside:
Check: \displaystyle -\frac{6}{18}(4 - 3x)^5 \cdot (-3) = (4 - 3x)^5. The minus sign from the inside derivative cancels the minus sign in front. When the inside is linear, as here, substitution is the rule “divide by the coefficient of x”.
Day 1: a constant factor can be adjusted, a variable factor cannot
(a) Find \int x^2\cos(x^3 + 1)\,dx. (b) Show that u = x^3 + 1 does not work for \int x\cos(x^3 + 1)\,dx.
Solution(a) The cosine holds x^3 + 1, whose derivative is 3x^2. Only x^2 is present, which differs by the constant factor 3. Let u = x^3 + 1, so du = 3x^2\,dx and \displaystyle x^2\,dx = \frac{1}{3}\,du:
Check: \displaystyle \frac{1}{3}\cos(x^3 + 1) \cdot 3x^2 = x^2\cos(x^3 + 1).
(b) Here du = 3x^2\,dx gives \displaystyle x\,dx = \frac{1}{3x}\,du, and the factor \displaystyle \frac{1}{3x} is not a constant, so the constant multiple rule does not let it move through the integral sign. The function that moving it would produce, \displaystyle \frac{1}{3x}\sin(x^3 + 1), shows directly why: by the product rule its derivative is
which has an extra term, so it is not an antiderivative of x\cos(x^3 + 1). Writing x = (u - 1)^{1/3} instead keeps the step honest but leaves \displaystyle \int \frac{\cos u}{3(u - 1)^{1/3}}\,du, which is no easier. The substitution works only when the leftover factor is a constant multiple of du.
Day 1: Yeng and Choua choose different inside functions
Find \displaystyle \int \frac{(\sqrt{x} + 1)^4}{\sqrt{x}}\,dx for x > 0. Yeng lets \displaystyle u = \sqrt{x} + 1; Choua lets \displaystyle u = \sqrt{x}.
SolutionYeng: \displaystyle du = \frac{1}{2\sqrt{x}}\,dx, so \displaystyle \frac{dx}{\sqrt{x}} = 2\,du. The integrand becomes 2u^4:
Choua: again \displaystyle \frac{dx}{\sqrt{x}} = 2\,du, and now \displaystyle \sqrt{x} + 1 = u + 1, so the integral is \int 2(u + 1)^4\,du. That is a power of the linear expression u + 1, which is \displaystyle \frac{2}{5}(u + 1)^5 + C (one more substitution, w = u + 1, or the rule for a linear inside). Putting \displaystyle u = \sqrt{x} back gives the same \displaystyle \frac{2}{5}(\sqrt{x} + 1)^5 + C.
Check: \displaystyle \frac{2}{5} \cdot 5(\sqrt{x} + 1)^4 \cdot \frac{1}{2\sqrt{x}} = \frac{(\sqrt{x} + 1)^4}{\sqrt{x}}. There is often more than one workable choice; the one that swallows the most of the integrand at once is usually the shortest.
Day 2: logarithms, trig and exponentials
Find (a) \displaystyle \int \frac{x}{x^2 + 5}\,dx, (b) \int \tan x\,dx, (c) \int \sin^4 x\cos x\,dx, (d) \displaystyle \int \frac{e^{1/x}}{x^2}\,dx.
Solution(a) The denominator x^2 + 5 has derivative 2x, and x is on top. Let u = x^2 + 5, so \displaystyle x\,dx = \frac{1}{2}\,du:
The bars of \ln\lvert u \rvert are dropped because x^2 + 5 > 0 for every x, so the answer holds on the whole real line.
(b) Write \displaystyle \tan x = \frac{\sin x}{\cos x}. The derivative of the denominator is -\sin x, so let u = \cos x and \sin x\,dx = -du:
since \displaystyle -\ln\lvert \cos x \rvert = \ln \frac{1}{\lvert \cos x \rvert}. It holds on each interval where \cos x \ne 0.
(c) Let u = \sin x, so du = \cos x\,dx and the integrand is u^4\,du:
(d) The exponent \displaystyle \frac{1}{x} has derivative \displaystyle -\frac{1}{x^2}. Let \displaystyle u = \frac{1}{x}, so \displaystyle \frac{dx}{x^2} = -du:
on each of x < 0 and x > 0. Check: \displaystyle -e^{1/x}\cdot\left(-\frac{1}{x^2}\right) = \frac{e^{1/x}}{x^2}.
Day 2: a stray x, and an integrand with no partner
(a) Find \displaystyle \int x\sqrt{x + 3}\,dx. (b) Try u = x^2 on \int \cos(x^2)\,dx.
Solution(a) Let u = x + 3, so du = dx. The root becomes u^{1/2}, but a factor x is left over. Solve the substitution for x: x = u - 3. Now every x is in terms of u:
Check: the derivative is (x + 3)^{3/2} - 3(x + 3)^{1/2}, which factors as (x + 3)^{1/2}\big((x + 3) - 3\big), that is, \displaystyle x\sqrt{x + 3}.
(b) With u = x^2, du = 2x\,dx, but there is no x in the integrand to pair with dx. Forcing it, for x > 0 we have \displaystyle x = \sqrt{u} and \displaystyle dx = \frac{du}{2\sqrt{u}}, so
which is no simpler. No other choice does better: by the theorem stated without proof on Day 2, \cos(x^2) has no antiderivative built from the usual functions. A single extra factor of x changes everything, which is why the factor must be checked for before a substitution is attempted.
Day 3: changing the limits, or changing back
(a) Evaluate \displaystyle \int_1^e \frac{(\ln x)^2}{x}\,dx both ways. (b) Evaluate \displaystyle \int_0^{\pi/2} \frac{\sin x}{(1 + \cos x)^2}\,dx.
(a) Let u = \ln x, so \displaystyle du = \frac{dx}{x}, and \displaystyle g'(x) = \frac{1}{x} is continuous on [1, e]. Changing the limits: x = 1 gives u = \ln 1 = 0, and x = e gives u = 1, so
Changing back: the antiderivative is \displaystyle \frac{u^3}{3} = \frac{(\ln x)^3}{3}, and
Figure 1 shows what the substitution did to the region: a long low region becomes a short tall one with the same area.
(b) The denominator holds 1 + \cos x, whose derivative -\sin x is on top up to sign. Let u = 1 + \cos x, so \sin x\,dx = -du. The limits: x = 0 gives u = 2, and \displaystyle x = \frac{\pi}{2} gives u = 1. The new lower limit is the larger one, and that is fine:
Swapping the limits absorbed the minus sign. The answer is positive, as it must be: the integrand is positive on \displaystyle \left(0, \frac{\pi}{2}\right].
Day 3: symmetry, and Choua’s pond
(a) Evaluate \int_{-\pi/2}^{\pi/2} \left(\sin^5 x + \cos x\right)dx.
(b) Choua fills a garden pond with a pump. Water enters at the rate \displaystyle r(t) = 12t\sqrt{t^2 + 16} liters per minute, t minutes after the pump starts, and the pond held 500 liters at t = 0. How much water enters in the first 3 minutes, and how much does the pond hold after T minutes, for 0 \le T \le 3?
Solution(a) An antiderivative of \sin^5 x needs the identities of Trigonometric integrals, but none is needed here. Since \sin(-x) = -\sin x, \sin^5(-x) = -\sin^5 x: the first term is odd and continuous, so it contributes 0 over the symmetric interval. The second term is even. So
(b) The water added is the integral of the rate. Let u = t^2 + 16, so du = 2t\,dt and 12t\,dt = 6\,du. The limits: t = 0 gives u = 16, and t = 3 gives u = 25:
For the amount at time T, the same substitution with upper limit u = T^2 + 16 gives
At T = 3 this is 244 + 500 = 744 liters, the starting 500 plus the 244 added. This is an accumulation function: by the first fundamental theorem, V'(T) = r(T).
Boundaries
- The substitution rule needs g' continuous and f continuous on the values g takes. Substitution does not rescue an integral whose integrand is unbounded on the interval: \displaystyle \int_0^2 \frac{2x}{x^2 - 1}\,dx does not exist, and with u = x^2 - 1 it becomes \displaystyle \int_{-1}^{3} \frac{du}{u}, which does not exist either.
- After the substitution, no x may remain, and x and u never share one integrand. Only a constant factor moves through an integral sign (Example 2).
- In a definite integral the limits belong to the variable of integration: u-limits with a u-integrand, x-limits with an x-integrand.
- \ln\lvert u \rvert + C is an antiderivative of \displaystyle \frac{1}{u} only on an interval where u \ne 0, so an answer such as \ln\lvert \sec x \rvert + C holds only on intervals where \cos x \ne 0.
- Solving for x needs care: from u = x^2, the step \displaystyle x = \sqrt{u} holds only for x \ge 0.
- Substitution finds an antiderivative only when the integrand is a chain-rule derivative up to a constant. Many integrands are not, and some, such as \cos(x^2), have no antiderivative built from the usual functions.
Where this goes next
Substitution reverses the chain rule; integration by parts reverses the product rule, and the two together handle most integrals met in a first course. Trigonometric substitution runs the idea in the other direction, setting x = \sin\theta or x = \tan\theta to clear a square root. Partial fractions splits a rational function into pieces that are each a logarithm or arctangent substitution, and separable differential equations integrate both sides, often by substitution, to solve an equation for a function. Choosing an integration technique gathers these into one way of reading an integrand.
Exercises
Day 1: one substitution
- Find \int 3x^2\left(x^3 + 2\right)^4 dx.
- Find \displaystyle \int \sqrt{5x + 1}\,dx.
- Find \int x\left(x^2 - 3\right)^5 dx.
- Find \int x^3\sin(x^4)\,dx.
- Find \displaystyle \int \frac{x^2}{\sqrt{x^3 + 8}}\,dx.
- Find \int (x + 1)\left(x^2 + 2x - 5\right)^3 dx.
Day 1: reasoning
- For \int 2x\left(x^2 + 1\right)dx, Yeng substitutes u = x^2 + 1 and gets \displaystyle \frac{\left(x^2 + 1\right)^2}{2} + C. Choua multiplies out and gets \displaystyle \frac{x^4}{2} + x^2 + C. Show that both are correct and explain how two different formulas can both be right.
- Ntxhi writes \displaystyle \int \cos(x^2)\,dx = \frac{\sin(x^2)}{2x} + C. Differentiate the answer to show it is wrong, and say which step was not allowed.
Day 2: logarithms, trig, exponentials
- Find \displaystyle \int \frac{x^3}{x^4 + 1}\,dx.
- Find \displaystyle \int \frac{2x + 3}{x^2 + 3x - 4}\,dx, and say why the absolute value bars are needed.
- Find \int \cot x\,dx.
- Find \int \cos^3 x\sin x\,dx.
- Find \displaystyle \int \frac{e^{\sqrt{x}}}{\sqrt{x}}\,dx.
- Find \displaystyle \int e^x\sqrt{e^x + 3}\,dx.
Day 2: a stray x
- Find \displaystyle \int \frac{x}{\sqrt{x + 1}}\,dx.
- Find \displaystyle \int x^3\sqrt{x^2 + 1}\,dx. (Write x^3\,dx = x^2 \cdot x\,dx.)
Day 2: reasoning
- For which of (a) \int x\sin(x^2 + 1)\,dx, (b) \int \sin(x^2 + 1)\,dx, (c) \int x^2\sin(x^2 + 1)\,dx does u = x^2 + 1 turn the integral into one in u alone? Evaluate that one, and say what is left over in the others.
Day 3: definite integrals
- Evaluate \displaystyle \int_0^2 \frac{x}{\left(x^2 + 1\right)^2}\,dx.
- Evaluate \displaystyle \int_0^{\pi/2} \cos x\sqrt{1 + \sin x}\,dx.
- Evaluate \displaystyle \int_0^{\ln 2} \frac{e^x}{\left(e^x + 1\right)^2}\,dx.
- Evaluate \displaystyle \int_2^5 x\sqrt{x - 1}\,dx.
- Evaluate \int_{-1}^{1} \left(x^4\sin x + 3x^2\right)dx.
- Find the area of the region under \displaystyle y = \frac{x}{\sqrt{x^2 + 9}} and above the x-axis, from x = 0 to x = 4.
Day 3: reasoning
- Ntxhi evaluates \int_0^1 2x\left(x^2 + 1\right)^3 dx with u = x^2 + 1 and writes \displaystyle \int_0^1 u^3\,du = \frac{1}{4}. Find the correct value and explain the error.
- Prove that \int_0^1 x^m(1 - x)^n\,dx = \int_0^1 x^n(1 - x)^m\,dx for all nonnegative integers m and n.
Answers
- \displaystyle \frac{\left(x^3 + 2\right)^5}{5} + C. With u = x^3 + 2, du = 3x^2\,dx exactly, and \displaystyle \int u^4\,du = \frac{u^5}{5} + C.
- \displaystyle \frac{2}{15}(5x + 1)^{3/2} + C. With u = 5x + 1, \displaystyle dx = \frac{1}{5}\,du, and \displaystyle \frac{1}{5}\cdot\frac{2}{3}u^{3/2} = \frac{2}{15}u^{3/2}.
- \displaystyle \frac{\left(x^2 - 3\right)^6}{12} + C. With u = x^2 - 3, \displaystyle x\,dx = \frac{1}{2}\,du, and \displaystyle \frac{1}{2}\cdot\frac{u^6}{6}.
- \displaystyle -\frac{1}{4}\cos(x^4) + C. With u = x^4, \displaystyle x^3\,dx = \frac{1}{4}\,du, and \displaystyle \frac{1}{4}\int \sin u\,du = -\frac{1}{4}\cos u + C.
- \displaystyle \frac{2}{3}\sqrt{x^3 + 8} + C. With u = x^3 + 8, \displaystyle x^2\,dx = \frac{1}{3}\,du, and \displaystyle \frac{1}{3}\int u^{-1/2}\,du = \frac{2}{3}u^{1/2} + C. It holds for x > -2.
- \displaystyle \frac{\left(x^2 + 2x - 5\right)^4}{8} + C. With u = x^2 + 2x - 5, du = (2x + 2)\,dx, so \displaystyle (x + 1)\,dx = \frac{1}{2}\,du and \displaystyle \frac{1}{2}\cdot\frac{u^4}{4}.
- Both differentiate to 2x^3 + 2x = 2x\left(x^2 + 1\right). Expanding Yeng’s answer gives \displaystyle \frac{x^4}{2} + x^2 + \frac{1}{2} + C, so the two formulas differ by the constant \displaystyle \frac{1}{2}. Two antiderivatives on an interval may differ by a constant, and the +\,C absorbs it.
- By the product rule the derivative is \displaystyle \cos(x^2) - \frac{\sin(x^2)}{2x^2}, not \cos(x^2). With u = x^2, \displaystyle dx = \frac{du}{2x}, and Ntxhi moved the non-constant factor \displaystyle \frac{1}{2x} outside the integral, which only a constant may do.
- \displaystyle \frac{1}{4}\ln\left(x^4 + 1\right) + C. With u = x^4 + 1 > 0, \displaystyle x^3\,dx = \frac{1}{4}\,du, and \displaystyle \frac{1}{4}\int \frac{du}{u}.
- \ln\lvert x^2 + 3x - 4 \rvert + C. With u = x^2 + 3x - 4, du = (2x + 3)\,dx exactly. Since x^2 + 3x - 4 = (x + 4)(x - 1) is negative for -4 < x < 1, the bars are needed there; the answer holds on each interval avoiding x = -4 and x = 1.
- \ln\lvert \sin x \rvert + C. Write \displaystyle \cot x = \frac{\cos x}{\sin x} and let u = \sin x, du = \cos x\,dx. It holds on each interval where \sin x \ne 0.
- \displaystyle -\frac{\cos^4 x}{4} + C. With u = \cos x, \sin x\,dx = -du, and \displaystyle -\int u^3\,du = -\frac{u^4}{4} + C.
- \displaystyle 2e^{\sqrt{x}} + C. With \displaystyle u = \sqrt{x}, \displaystyle du = \frac{dx}{2\sqrt{x}}, so \displaystyle \frac{dx}{\sqrt{x}} = 2\,du and 2\int e^u\,du.
- \displaystyle \frac{2}{3}\left(e^x + 3\right)^{3/2} + C. With u = e^x + 3, du = e^x\,dx exactly, and \displaystyle \int u^{1/2}\,du = \frac{2}{3}u^{3/2} + C.
- \displaystyle \frac{2}{3}(x + 1)^{3/2} - 2(x + 1)^{1/2} + C. With u = x + 1 and x = u - 1, the integral is \int \left(u^{1/2} - u^{-1/2}\right)du, which is \displaystyle \frac{2}{3}u^{3/2} - 2u^{1/2} + C, for x > -1.
- \displaystyle \frac{1}{5}\left(x^2 + 1\right)^{5/2} \displaystyle - \frac{1}{3}\left(x^2 + 1\right)^{3/2} + C. With u = x^2 + 1, \displaystyle x\,dx = \frac{1}{2}\,du and x^2 = u - 1, so the integral is \displaystyle \frac{1}{2}\int \left(u^{3/2} - u^{1/2}\right)du, which is \displaystyle \frac{1}{2}\left(\frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2}\right) + C.
- Only (a): \displaystyle -\frac{1}{2}\cos(x^2 + 1) + C, since \displaystyle x\,dx = \frac{1}{2}\,du and \displaystyle \frac{1}{2}\int \sin u\,du = -\frac{1}{2}\cos u + C. In (b), \displaystyle dx = \frac{du}{2x} leaves a factor \displaystyle \frac{1}{x}; in (c), x^2\,dx = x \cdot x\,dx leaves a factor \displaystyle x = \sqrt{u - 1} (for x \ge 0), and the integral \displaystyle \frac{1}{2}\int \sqrt{u - 1}\,\sin u\,du is no simpler.
- \displaystyle \frac{2}{5}. With u = x^2 + 1, the limits become 1 and 5, and \displaystyle \frac{1}{2}\int_1^5 u^{-2}\,du is \displaystyle \frac{1}{2}\left(-\frac{1}{u}\right)\Big|_1^5 = \frac{1}{2}\left(1 - \frac{1}{5}\right).
- \displaystyle \frac{4\sqrt{2} - 2}{3}. With u = 1 + \sin x, the limits become 1 and 2, and \displaystyle \frac{2}{3}u^{3/2}\Big|_1^2 = \frac{2}{3}\left(2\sqrt{2} - 1\right).
- \displaystyle \frac{1}{6}. With u = e^x + 1, the limits become 2 and 3, and \displaystyle -\frac{1}{u}\Big|_2^3 = -\frac{1}{3} + \frac{1}{2}.
- \displaystyle \frac{256}{15}. With u = x - 1 and x = u + 1, the limits become 1 and 4. The antiderivative \displaystyle \frac{2}{5}u^{5/2} + \frac{2}{3}u^{3/2} gives \displaystyle \frac{272}{15} - \frac{16}{15}.
- 2. x^4\sin x is odd (an even factor times an odd one) and continuous, so it gives 0; 3x^2 is even, and 2\int_0^1 3x^2\,dx = 2.
- 2. The integrand is positive for x > 0. With u = x^2 + 9, the limits become 9 and 25, and \displaystyle \frac{1}{2}\int_9^{25} u^{-1/2}\,du = \sqrt{u}\Big|_9^{25} = 5 - 3.
- \displaystyle \frac{15}{4}. The limits 0 and 1 are values of x, but the integral is in u. At x = 0, u = 1, and at x = 1, u = 2, so the integral is \displaystyle \int_1^2 u^3\,du = \frac{16 - 1}{4}. Back-substituting agrees: \displaystyle \frac{\left(x^2 + 1\right)^4}{4}\Big|_0^1 = 4 - \frac{1}{4}.
- Let u = 1 - x, so du = -dx and x = 1 - u; x = 0 gives u = 1 and x = 1 gives u = 0. Then \int_0^1 x^m(1 - x)^n\,dx becomes \int_1^0 (1 - u)^m u^n\,(-du), which is \int_0^1 u^n(1 - u)^m\,du, and renaming the variable u as x does not change the value.