Here’s a simple integral that we can’t yet evaluate:
∫
xcosx dx.
It’s a simple ma er to take the deriva ve of the integrand using the Product Rule, but there is no Product Rule for integrals. However, this sec on introduces Integra on by Parts, a method of integra on that is based on the Product Rule for deriva ves. It will enable us to evaluate this integral.
The Product Rule says that ifuandvare func ons ofx, then(uv)′ =u′v+uv′. For simplicity, we’ve wri enuforu(x)andvforv(x). Suppose we integrate both sides with respect tox. This gives
∫
(uv)′dx=
∫
(u′v+uv′)dx.
By the Fundamental Theorem of Calculus, the le side integrates touv. The right side can be broken up into two integrals, and we have
uv=
∫
u′v dx+
∫ uv′dx.
Solving for the second integral we have
∫
uv′dx=uv−
∫ u′v dx.
Using differen al nota on, we can writedu = u′(x)dxanddv = v′(x)dxand the expression above can be wri en as follows:
∫
u dv=uv−
∫ v du.
This is the Integra on by Parts formula. For reference purposes, we state this in a theorem.
Theorem 48 Integra on by Parts
Letuandvbe differen able func ons ofxon an intervalIcontaininga
andb. Then ∫
u dv=uv−
∫ v du,
and ∫ x=b
x=a
u dv=uvb
a−
∫ x=b x=a
v du.
Notes:
Let’s try an example to understand our new technique.
Example 159 Integra ng using Integra on by Parts Evaluate
∫
xcosx dx.
S The key to Integra on by Parts is to iden fy part of the in- tegrand as “u” and part as “dv.” Regular prac ce will help one make good iden- fica ons, and later we will introduce some principles that help. For now, let u=xanddv=cosx dx.
It is generally useful to make a small table of these values as done below.
Right now we only knowuanddvas shown on the le of Figure 6.3; on the right we fill in the rest of what we need. Ifu=x, thendu=dx. Sincedv=cosx dx, vis an an deriva ve of cosx. We choosev=sinx.
u=x v=?
du=? dv=cosx dx ⇒ u=x v=sinx
du=dx dv=cosx dx Figure 6.3: Se ng up Integra on by Parts.
Now subs tute all of this into the Integra on by Parts formula, giving
∫
xcosx dx=xsinx−
∫
sinx dx.
We can then integrate sinxto get−cosx+Cand overall our answer is
∫
xcosx dx=xsinx+cosx+C.
Note how the an deriva ve contains a product,xsinx. This product is what makes Integra on by Parts necessary.
The example above demonstrates how Integra on by Parts works in general.
We try to iden fyuanddvin the integral we are given, and the key is that we usually want to chooseuanddvso thatduis simpler thanuandvis hopefully not too much more complicated thandv. This will mean that the integral on the right side of the Integra on by Parts formula,∫
v duwill be simpler to integrate than the original integral∫
u dv.
In the example above, we choseu=xanddv=cosx dx. Thendu=dxwas simpler thanuandv=sinxis no more complicated thandv. Therefore, instead of integra ngxcosx dx, we could integrate sinx dx, which we knew how to do.
A useful mnemonic for helping to determineuis “LIATE,” where L =Logarithmic, I =Inverse Trig., A =Algebraic (polynomials),
T =Trigonometric, and E =Exponen al.
Notes:
If the integrand contains both a logarithmic and an algebraic term, in general le ngube the logarithmic term works best, as indicated by L coming before A in LIATE.
We now consider another example.
Example 160 Integra ng using Integra on by Parts Evaluate
∫ xexdx.
S The integrand contains anAlgebraic term (x) and anExponen al term (ex). Our mnemonic suggests le ngube the algebraic term, so we choose u=xanddv=exdx. Thendu=dxandv=exas indicated by the tables below.
u=x v=?
du=? dv=exdx ⇒ u=x v=ex
du=dx dv=exdx Figure 6.4: Se ng up Integra on by Parts.
We seeduis simpler thanu, while there is no change in going fromdvtov.
This is good. The Integra on by Parts formula gives
∫
xexdx=xex−
∫ exdx.
The integral on the right is simple; our final answer is
∫
xexdx=xex−ex+C.
Note again how the an deriva ves contain a product term.
Example 161 Integra ng using Integra on by Parts Evaluate
∫
x2cosx dx.
S The mnemonic suggests le ngu=x2instead of the trigono- metric func on, hencedv=cosx dx. Thendu=2x dxandv=sinxas shown below.
u=x2 v=?
du=? dv=cosx dx ⇒ u=x2 v=sinx
du=2x dx dv=cosx dx Figure 6.5: Se ng up Integra on by Parts.
Notes:
The Integra on by Parts formula gives
∫
x2cosx dx=x2sinx−
∫
2xsinx dx.
At this point, the integral on the right is indeed simpler than the one we started with, but to evaluate it, we need to do Integra on by Parts again. Here we chooseu=2xanddv=sinxand fill in the rest below.
u=2x v=?
du=? dv=sinx dx ⇒ u=2x v=−cosx
du=2dx dv=sinx dx Figure 6.6: Se ng up Integra on by Parts (again).
∫
x2cosx dx=x2sinx− (
−2xcosx−
∫
−2 cosx dx )
.
The integral all the way on the right is now something we can evaluate. It eval- uates to−2 sinx. Then going through and simplifying, being careful to keep all the signs straight, our answer is
∫
x2cosx dx=x2sinx+2xcosx−2 sinx+C.
Example 162 Integra ng using Integra on by Parts Evaluate
∫
excosx dx.
S This is a classic problem. Our mnemonic suggests le ngu be the trigonometric func on instead of the exponen al. In this par cular ex- ample, one can letube either cosxorex; to demonstrate that we do not have to follow LIATE, we chooseu =exand hencedv =cosx dx. Thendu =exdx andv=sinxas shown below.
u=ex v=?
du=? dv=cosx dx ⇒ u=ex v=sinx
du=exdx dv=cosx dx Figure 6.7: Se ng up Integra on by Parts.
No ce thatduis no simpler thanu, going against our general rule (but bear with us). The Integra on by Parts formula yields
∫
excosx dx=exsinx−
∫
exsinx dx.
Notes:
The integral on the right is not much different than the one we started with, so it seems like we have go en nowhere. Let’s keep working and apply Integra on by Parts to the new integral, usingu=exanddv=sinx dx. This leads us to the following:
u=ex v=?
du=? dv=sinx dx ⇒ u=ex v=−cosx du=exdx dv=sinx dx Figure 6.8: Se ng up Integra on by Parts (again).
The Integra on by Parts formula then gives:
∫
excosx dx=exsinx− (
−excosx−
∫
−excosx dx )
=exsinx+excosx−
∫
excosx dx.
It seems we are back right where we started, as the right hand side contains
∫excosx dx. But this is actually a good thing.
Add
∫
excosx dxto both sides. This gives
2
∫
excosx dx=exsinx+excosx Now divide both sides by 2:
∫
excosx dx=1 2
(exsinx+excosx) .
Simplifying a li le and adding the constant of integra on, our answer is thus
∫
excosx dx= 1
2ex(sinx+cosx) +C.
Example 163 Integra ng using Integra on by Parts: an deriva ve oflnx Evaluate
∫ lnx dx.
S One may have no ced that we have rules for integra ng the familiar trigonometric func ons andex, but we have not yet given a rule for integra ng lnx. That is because lnxcan’t easily be integrated with any of the rules we have learned up to this point. But we can find its an deriva ve by a
Notes:
clever applica on of Integra on by Parts. Setu= lnxanddv = dx. This is a good, sneaky trick to learn as it can help in other situa ons. This determines du= (1/x)dxandv=xas shown below.
u=lnx v=?
du=? dv=dx ⇒ u=lnx v=x
du=1/x dx dv=dx Figure 6.9: Se ng up Integra on by Parts.
Pu ng this all together in the Integra on by Parts formula, things work out
very nicely: ∫
lnx dx=xlnx−
∫ x1
x dx.
The new integral simplifies to∫
1dx, which is about as simple as things get. Its integral isx+Cand our answer is
∫
lnx dx=xlnx−x+C.
Example 164 Integra ng using Int. by Parts: an deriva ve ofarctanx Evaluate
∫
arctanx dx.
S The same sneaky trick we used above works here. Letu= arctanxanddv=dx. Thendu=1/(1+x2)dxandv=x. The Integra on by Parts formula gives
∫
arctanx dx=xarctanx−
∫ x 1+x2dx.
The integral on the right can be solved by subs tu on. Takingu=1+x2, we getdu=2x dx. The integral then becomes
∫
arctanx dx=xarctanx−1 2
∫ 1 udu.
The integral on the right evaluates to ln|u|+C, which becomes ln(1+x2) +C.
Therefore, the answer is
∫
arctanx dx=xarctanx−ln(1+x2) +C.
Notes:
Subs tu on Before Integra on
When taking deriva ves, it was common to employ mul ple rules (such as using both the Quo ent and the Chain Rules). It should then come as no surprise that some integrals are best evaluated by combining integra on techniques. In par cular, here we illustrate making an “unusual” subs tu on first before using Integra on by Parts.
Example 165 Integra on by Parts a er subs tu on Evaluate
∫
cos(lnx)dx.
S The integrand contains a composi on of func ons, leading us to think Subs tu on would be beneficial. Le ngu = lnx, we havedu = 1/x dx. This seems problema c, as we do not have a 1/xin the integrand. But consider:
du=1
x dx⇒x·du=dx.
Sinceu=lnx, we can use inverse func ons and conclude thatx=eu. Therefore we have that
dx=x·du
=eudu.
We can thus replace lnxwithuanddxwitheudu. Thus we rewrite our integral
as ∫
cos(lnx)dx=
∫
eucosu du.
We evaluated this integral in Example 162. Using the result there, we have:
∫
cos(lnx)dx=
∫
eucosu du
= 1 2eu(
sinu+cosu) +C
= 1 2elnx(
sin(lnx) +cos(lnx)) +C
= 1 2x(
sin(lnx) +cos(lnx)) +C.
Definite Integrals and Integra on By Parts
So far we have focused only on evalua ng indefinite integrals. Of course, we can use Integra on by Parts to evaluate definite integrals as well, as Theorem
Notes:
48 states. We do so in the next example.
Example 166 Definite integra on using Integra on by Parts Evaluate
∫ 2 1
x2lnx dx.
S Our mnemonic suggests le ngu=lnx, hencedv =x2dx.
We then getdu= (1/x)dxandv=x3/3 as shown below.
u=lnx v=?
du=? dv=x2dx ⇒ u=lnx v=x3/3
du=1/x dx dv=x2dx Figure 6.10: Se ng up Integra on by Parts.
The Integra on by Parts formula then gives
∫ 2 1
x2lnx dx= x3 3 lnx
2
1
−
∫ 2 1
x3 3
1 xdx
= x3 3 lnx
2
1
−
∫ 2 1
x2 3 dx
= x3 3 lnx
2
1
−x3 9
2
1
= (x3
3 lnx−x3 9
) 2
1
= (8
3ln 2−8 9 )
− (1
3ln 1−1 9 )
= 8
3ln 2−7 9
≈1.07.
In general, Integra on by Parts is useful for integra ng certain products of func ons, like∫
xexdxor∫
x3sinx dx. It is also useful for integrals involving logarithms and inverse trigonometric func ons.
As stated before, integra on is generally more difficult than deriva on. We are developing tools for handling a large array of integrals, and experience will tell us when one tool is preferable/necessary over another. For instance, con- sider the three similar–looking integrals
∫ xexdx,
∫
xex2dx and
∫
xex3dx.
Notes:
While the first is calculated easily with Integra on by Parts, the second is best approached with Subs tu on. Taking things one step further, the third integral has no answer in terms of elementary func ons, so none of the methods we learn in calculus will get us the exact answer.
Integra on by Parts is a very useful method, second only to subs tu on. In the following sec ons of this chapter, we con nue to learn other integra on techniques. The next sec on focuses on handling integrals containing trigono- metric func ons.
Notes:
Terms and Concepts
1. T/F: Integra on by Parts is useful in evalua ng integrands that contain products of func ons.
2. T/F: Integra on by Parts can be thought of as the “opposite of the Chain Rule.”
3. For what is “LIATE” useful?
Problems
In Exercises 4 – 33, evaluate the given indefinite integral.
4.
∫
xsinx dx
5.
∫ xe−xdx
6.
∫
x2sinx dx
7.
∫
x3sinx dx
8.
∫ xex2dx
9.
∫ x3exdx
10.
∫
xe−2xdx
11.
∫
exsinx dx
12.
∫
e2xcosx dx
13.
∫
e2xsin(3x)dx
14.
∫
e5xcos(5x)dx
15.
∫
sinxcosx dx
16.
∫
sin−1x dx
17.
∫
tan−1(2x)dx
19. sin−1x dx
20.
∫ xlnx dx
21.
∫
(x−2)lnx dx
22.
∫
xln(x−1)dx
23.
∫
xln(x2)dx
24.
∫
x2lnx dx
25.
∫
(lnx)2dx
26.
∫
(ln(x+1))2 dx
27.
∫
xsec2x dx
28.
∫
xcsc2x dx
29.
∫ x√
x−2dx
30.
∫ x√
x2−2dx
31.
∫
secxtanx dx
32.
∫
xsecxtanx dx
33.
∫
xcscxcotx dx
In Exercises 34 – 38, evaluate the indefinite integral a er first making a subs tu on.
34.
∫
sin(lnx)dx
35.
∫ sin(√
x)dx
36.
∫ ln(√
x)dx
38. e dx
In Exercises 39 – 47, evaluate the definite integral. Note: the corresponding indefinite integrals appear in Exercises 4 – 12.
39.
∫ π 0
xsinx dx
40.
∫ 1
−1xe−xdx
41.
∫ π/4
−π/4
x2sinx dx
42.
∫ π/2
−π/2
x3sinx dx
43.
0
xex dx
44.
∫1
0
x3exdx
45.
∫2
1
xe−2xdx
46.
∫ π 0
exsinx dx
47.
∫ π/2
−π/2
e2xcosx dx