Integrating f(ax + b)

A stretched input costs a divide.

Why a divide appears

Differentiate sin 2x by the chain rule: the outside gives cos 2x, and the inside 2x has derivative 2, so the derivative of sin 2x is 2 cos 2x. The chain rule has brought out a factor of 2.

Integration runs this backwards, so it must take that factor away again. The integral of cos 2x is not sin 2x, which differentiates to twice too much, but half of it: ∫ cos 2x dx = ½ sin 2x + c. Check: the derivative of ½ sin 2x is ½ × 2 cos 2x = cos 2x.

The rule

Let F be an antiderivative of f. Then ∫ f(ax + b) dx = F(ax + b)/a + c. Integrate as though the inside were plain x, then divide by a, the coefficient of x inside.

It is the substitution u = ax + b, done once for every linear inside. Then du = a dx, so dx = du/a, and the integral becomes 1/a times the integral of f(u) du, which is F(u)/a + c. Check by the chain rule: F(ax + b) differentiates to f(ax + b) × a, and dividing by a leaves f(ax + b).

The constant b changes nothing. It slides the graph sideways, and its derivative is 0, so the chain rule brings out no factor for it.

xy

The gold curve y = cos 2x is the dashed curve y = cos x squeezed toward the y-axis by a factor of 2: it reaches the axis at π/4, the first dot, instead of π/2, the second. Every width is halved, so the shaded area, ½ sin(π/2) = ½, is half the area 1 under the dashed curve from 0 to π/2.

Powers of a bracket

For ∫ (3x + 1)⁴ dx, integrate as though the bracket were x: raise the power to 5 and divide by 5. Then divide by the inside coefficient 3. The answer is (3x + 1)⁵/15 + c. Check: 5(3x + 1)⁴ × 3 / 15 = (3x + 1)⁴.

For ∫ (2x + 1)³ dx, raise the power to 4, divide by 4 and then by 2: (2x + 1)⁴/8 + c. Check: 4(2x + 1)³ × 2 / 8 = (2x + 1)³.

The same divide works for any outer function. ∫ e^(5x) dx = e^(5x)/5 + c, because e^(5x) differentiates to 5e^(5x). A negative coefficient divides by a negative number: ∫ (5 − 2x)³ dx = −(5 − 2x)⁴/8 + c. Check: −4(5 − 2x)³ × (−2) / 8 = (5 − 2x)³.

A definite integral needs nothing new. The integral of (2x + 1)³ from 0 to 1 is 3⁴/8 − 1⁴/8 = 80/8 = 10, and 1000 midpoint strips add to 9.999998.

Only for a linear inside

The shortcut divides by the derivative of the inside, which for ax + b is the constant a. A curved inside has a derivative that changes with x, and dividing by it does not undo the chain rule.

Try it on (x² + 1)², whose inside x² + 1 has derivative 2x. Dividing by 2x would give (x² + 1)³/(6x). At x = 1 that expression has gradient 8/3, but the integrand there is (1 + 1)² = 4, so the shortcut fails. Multiplying out instead, (x² + 1)² = x⁴ + 2x² + 1, which integrates to x⁵/5 + 2x³/3 + x + c, and its gradient at x = 1 is 1 + 2 + 1 = 4.

The usual mistakes

Leaving out the divide. sin 2x differentiates to 2 cos 2x, twice the integrand cos 2x.

Multiplying by a instead of dividing. 2 sin 2x differentiates to 4 cos 2x, and 5e^(5x) to 25e^(5x).

Dividing by the new power only. (2x + 1)⁴/4 differentiates to 2(2x + 1)³; it needs the further divide by 2.

Dividing by b. The constant b only slides the input along, so the extra divide is by a.

A seismograph’s pulse

In the application below, the ground’s velocity is a sine whose inside is a straight line in time. Integrating it gives a cosine divided by the inside coefficient, and a second, narrower pulse has a larger coefficient and so a larger divisor.

Worked example: A Seismograph's Delayed Pulse: An Integral Whose Inside Is a Straight Line in Time

Question A seismograph records the ground's velocity. The first pulse arrives three seconds after the trigger and is recorded as v = 18sin(π(n − 3)4) millimeters per second for 3 ≤ n ≤ 7, where n is the number of seconds after the trigger. A smaller second pulse follows: v = 6sin(π(n − 9)2) millimeters per second for 9 ≤ n ≤ 11; each record reads 0 at both ends of its own pulse, and the ground is still between the pulses. The ground's displacement over a pulse is the integral of its velocity. (a) How far does the ground move during the first pulse? (b) How far during the second, and how far in all? Give each answer to three significant figures.

  1. 1.The inside of the first sine is π(n−3)4, a straight line in n whose coefficient of n is π4. Integrating a sine gives minus a cosine, and the whole result is then divided by that coefficient.

    061218357911n, seconds after the triggervelocity, mm/sinside the first sine: pi(n − 3)/4
    061218357911n, seconds after the triggervelocity, mm/sinside the first sine: pi(n − 3)/4
    The inside, π(n−3)4, is a straight line in n whose coefficient of n is π4.

  2. 2.So ∫ 18sin(π(n−3)4)dn = −18 × 4πcos(π(n−3)4) + C = −72πcos(π(n−3)4) + C.

    061218357911n, seconds after the triggervelocity, mm/sinside the first sine: pi(n − 3)/4integral = −72/pi × cos(pi(n − 3)/4) + C
    061218357911n, seconds after the triggervelocity, mm/sinside the first sine: pi(n − 3)/4integral = −72/pi × cos(pi(n − 3)/4) + C
    Integrating the sine and dividing by that coefficient gives −72πcos(π(n−3)4) + C.

  3. 3.(a) At n = 7 the inside is π and the cosine is −1; at n = 3 the inside is 0 and the cosine is 1. The displacement is 72π − (−72π) = 144π = 45.8 millimeters.

    061218357911n, seconds after the triggervelocity, mm/s45.8 mminside the first sine: pi(n − 3)/4integral = −72/pi × cos(pi(n − 3)/4) + C(a) 72/pi + 72/pi = 144/pi = 45.8 mm
    061218357911n, seconds after the triggervelocity, mm/s45.8 mminside the first sine: pi(n − 3)/4integral = −72/pi × cos(pi(n − 3)/4) + C(a) 72/pi + 72/pi = 144/pi = 45.8 mm
    (a) The cosine runs from 1 to −1, so the shaded area is 144π = 45.8 millimeters.

  4. 4.The second pulse has π2 inside, twice as large, so the divisor is twice as large too: ∫ 6sin(π(n−9)2)dn = −6 × 2πcos(π(n−9)2) + C = −12πcos(π(n−9)2) + C.

    061218357911n, seconds after the triggervelocity, mm/s45.8 mminside the first sine: pi(n − 3)/4integral = −72/pi × cos(pi(n − 3)/4) + C(a) 72/pi + 72/pi = 144/pi = 45.8 mmsecond pulse: divide by pi/2, not pi/4
    061218357911n, seconds after the triggervelocity, mm/s45.8 mminside the first sine: pi(n − 3)/4integral = −72/pi × cos(pi(n − 3)/4) + C(a) 72/pi + 72/pi = 144/pi = 45.8 mmsecond pulse: divide by pi/2, not pi/4
    The second pulse has π2 inside, so the divisor is twice as large and the antiderivative is −12πcos(π(n−9)2) + C.

  5. 5.(b) At n = 11 the inside is π and at n = 9 it is 0, so the second displacement is 12π + 12π = 24π = 7.64 millimeters, and the ground moves 144π + 24π = 168π = 53.5 millimeters in all. Check: the second pulse is a third of the height and half the width of the first, so a sixth of the displacement, and 45.8 ÷ 6 = 7.64.

    061218357911n, seconds after the triggervelocity, mm/s45.8 mm7.64 mm53.5 mm in allinside the first sine: pi(n − 3)/4integral = −72/pi × cos(pi(n − 3)/4) + C(a) 72/pi + 72/pi = 144/pi = 45.8 mmsecond pulse: divide by pi/2, not pi/4(b) 24/pi = 7.64 mm, and 168/pi = 53.5 mm in all
    061218357911n, seconds after the triggervelocity, mm/s45.8 mm7.64 mm53.5 mm in allinside the first sine: pi(n − 3)/4integral = −72/pi × cos(pi(n − 3)/4) + C(a) 72/pi + 72/pi = 144/pi = 45.8 mmsecond pulse: divide by pi/2, not pi/4(b) 24/pi = 7.64 mm, and 168/pi = 53.5 mm in all
    (b) The second pulse gives 24π = 7.64 millimeters, so the ground moves 168π = 53.5 millimeters in all.

Answer: (a) 144π = 45.8 millimeters; (b) 24π = 7.64 millimeters from the second pulse, and 168π = 53.5 millimeters in all

Common mistakes

  • Forgetting the divisor and writing the first displacement as 18 × 2 = 36 millimeters. Differentiating −18cos(π(n−3)4) gives 18π4sin(π(n−3)4), which is π4 times too large, so the antiderivative has to carry the 4π.
  • Rewriting the integrand in the new variable but keeping the old limits. Substituting u = π(n−3)4 turns the integral into 72π∫sin u du, and the limits must travel with it, from u = 0 to u = π. Keeping 3 and 7 alongside an integrand in u puts two variables in one integral, and the answer means nothing.

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