Kenneth Kuttler

196 CHAPTER 8. METHODS FOR FINDING ANTIDERIVATIVES

= xarctan(x)− 12

∫ 2x1+ x2 dx = xarctan(x)− 1

2ln(1+ x2)+C.

This trick works for arctan, ln, and various other inverse trig. functions.Sometimes you want to find antiderivatives for something like

∫f gdx where f (m) = 0

for some positive integer m. For example,∫

x5 sinxdx. If you do integration by parts re-peatedly, what do you get? Let G′

1 = g,G′2 = G1,G′

3 = G2 etc. Then the first applica-tion of integration by parts yields f G1 −

∫G1 f ′dx. The next application of integration by

parts yields f G1 −G2 f ′+∫

G2 f ′′dx. Yet another application of integration by parts yieldsf G1 −G2 f ′+G3 f ′′−

∫G3 f ′′′dx. Eventually the process will stop because a high enough

derivative of f equals zero. This justifies the following procedure for finding antiderivativesin this case.

Procedure 8.3.5 Suppose f (m) = 0 for some m a positive integer and let G′k = Gk−1

for all k and G0 = g. Then∫f gdx = f G1 − f ′G2 + f ′′G3 − f ′′′G4 + · · ·

Just keep writing these terms, alternating signs until the process yields a zero. Then add onan arbitrary constant of integration and stop. Sometimes people remember this in the formof a table.

f +→ G1

f ′ −→ G2

f ′′ +→ G3

f ′′′ −→ G4

Thus you fill in the table until the left column ends in a 0 and then do the arrows, f G1 −f ′G2 + f ′′G3 · · · till the process ends. Then add C, a constant of integration.

Example 8.3.6 Find∫

x5 sinxdx.

From the above procedure, and letting f (x) = x5, this equals

x5 (−cos(x))−5x4 (−sin(x))+20x3 (cos(x))−60x2 (sin(x))+120x(−cos(x))−120(−sin(x))+C.

To determine the distance an object moves for t ∈ [a,b] , one computes∫ b

a |v(t)|dt. Thereason is as follows. If v(t) is nonnegative on an interval [c,d], this means the position isincreasing. To find the distance travelled, you would consider r (d)− r (c) =

∫ ba |v(t)|dt.

If v(t) < 0, on an interval [c,d] this means r (t) is decreasing. Thus the distance on thisinterval is r (c)− r (d) which equals

∫ dc −v(t)dt =

∫ dc |v(t)|dt. Splitting the interval into

sub-intervals on which the velocity is either positive or negative, one obtains that the dis-tance traveled is

∫ ba |v(t)|dt. This motivates the following definition.

Definition 8.3.7 Let the position of an object moving on R be denoted as r (t) . Thenthe distance moved for t ∈ [a,b] is

∫ ba |r′ (t)|dt.

196 CHAPTER 8. METHODS FOR FINDING ANTIDERIVATIVES1 2x2/ 14xThis trick works for arctan, In, and various other inverse trig. functions.Sometimes you want to find antiderivatives for something like [{ fgdx where f (™) — 9for some positive integer m. For example, [x sinxdx. If you do integration by parts re-peatedly, what do you get? Let G = g,G) = G,,G4 = G etc. Then the first applica-tion of integration by parts yields fG, — { G,f’dx. The next application of integration byparts yields fG, — Gof’ + f Gof”dx. Yet another application of integration by parts yields{Gi —Gof'+ G3f" — f G3f'"dx. Eventually the process will stop because a high enoughderivative of f equals zero. This justifies the following procedure for finding antiderivativesin this case.1= xarctan (x) dx = xarctan (x) — zn (1+) +C.Procedure 8.3.5 Suppose f ") =0 for some ma positive integer and let Gi. = Ge-1for all k and Go = g. Then[ feax= se; — f’Gy + f"G3— f" Gate:Just keep writing these terms, alternating signs until the process yields a zero. Then add onan arbitrary constant of integration and stop. Sometimes people remember this in the formof a table.&f > Gf 4 Gf" 4+ Gf" + GsThus you fill in the table until the left column ends in a 0 and then do the arrows, fG, —f'Go+ f"G3--- till the process ends. Then add C, a constant of integration.Example 8.3.6 Find { x sinxdx.From the above procedure, and letting f (x) = x°, this equalsx (—cos (x)) — 5x4 (—sin (x)) + 20x (cos (x)) — 60x? (sin (x))+120x(—cos(x)) — 120(—sin(x)) +C.To determine the distance an object moves for t € [a,b], one computes ? |v (t)|dt. Thereason is as follows. If v(t) is nonnegative on an interval [c,d], this means the position isincreasing. To find the distance travelled, you would consider r(d) —r(c) = f° |v (t)| dt.If v(t) <0, on an interval [c,d] this means r(t) is decreasing. Thus the distance on thisinterval is r(c) — r(d) which equals fd —v(t)dt = sé |v(t)|dt. Splitting the interval intosub-intervals on which the velocity is either positive or negative, one obtains that the dis-tance traveled is [ iM |v(t)| dt. This motivates the following definition.Definition 8.3.7 Let the position of an object moving on R be denoted as r(t). Thenthe distance moved for t € [a,b] is f? |r’ (t)| dt.