MA 16021
Integration by parts summary and examples
K. Rotz
Like derivatives, integrals of products are not well-behaved. In general,
Z
Z
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f (x)g(x) dx 6= ( f (x) dx)( g(x) dx).
Almost1 always when you have to integrate a product (or quotient) of functions, you have to use one
of the two tools that we talked about: u-substitution or integration by parts. Since u-substitution
is based on the chain rule
derivatives,
it is most useful when the integrand has a function within
R for
R
2
another function, e.g. xex dx or sin x sin(cos x) dx. Integration by parts is most useful whenever
u-substitution fails.
Once you decide to use integration by parts, the basic steps for performing the integration are as
follows.
1. Choose u and dv: To do this, look at the factors of the integrand and classify them as either
being Logarithmic, Algebraic, Trigonometric, or Exponential. The first one that appears in the
acronym LATE is what you should choose to be your u. The remaining stuff in the integral is
what you choose for dv.
2. Find du and v based on step 1: Differentiate u to find du (don’t forget about the chain rule!)
and integrate dv to find v (you may need to do a u-sub to find v).
3. Apply the formula: The integration by parts formula is
Z
Z
u dv = uv −
v du .
Dump the results of the first two steps into the formula and simplify.
At this point there are a few possibilities:
R
(a) You can perform the integration v du either by using the power rule, exponential rule, a
trig integral, x1 rule, or by doing a u-substitution.
(b) You can’t do the Rintegral because it’s even harder
to doR than the original. For example, if
R
you started with x sin x dx, then if you got v du = x2 cos x dx, that’s even harder to
do than the original integral. In this case, something’s probably messed up in the first few
steps, so go back and try to trace your error.
1
R
There are very few exceptions to this. One exception,
for example, would be sec x tan x dx. We know that the
R
derivative of sec x is sec x tan x, so you can just right to sec x tan x dx = sec x + C.
1
MA 16021
Integration by parts summary and examples
K. Rotz
R
(c) You might have to integrate by parts again in order to compute v du. This happens a
lot
you’reR integrating powers of x times trig functions
or exponential functions, e.g.
R 2when
R
−2x
3
xe
dx or x sin x dx. In the former case, the v du part after integrating by parts
once is
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xe−2x dx.
You still can’t just do the integral using u-sub, but notice that the power on the x has lowered
by one. If you integrate by parts
the Rpower of x will lower by 1 again (giving x0 = 1),
R 0 again,
and you can easily integrate x e−2x dx = e−2x dx.
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√
Example 1. Find
x ln(x2 ) dx.
Solution: First, you should convince yourself that integration by parts is necessary. Since there’s a
product, we either have to do u-sub or parts. The second factor does indeed involve a composition, so
we could try u-sub with u = x2 (the stuff inside of the √
composition). However, du = 2x dx in that case,
and that doesn’t fit with the rest of the integral (the x dx part). Therefore u-sub probably isn’t the
way to proceed. Parts it is!
√
1. Between the two factors, x is Algebraic and ln(x2 ) is Logarithmic. LATE then tells us that
u = ln x is the best choice. The leftover stuff is what we pick for dv.
u = ln(x2 ),
dv = x1/2 dx.
2. Differentiate u (don’t forget the chain rule!) and integrate dv (use the power rule):
1
2
2
v = x3/2
du = ( 2 )(2x) dx = dx,
x
x
3
3. Apply the IBP formula:
uv
v
du
z
}|
{ Z z }| { z }| {
√
2
2
2
x ln(x2 ) dx = x3/2 ln(x2 ) − ( x3/2 ) ( ) dx
3
3
x
Z
4
2 3/2
x1/2 dx.
= x ln(x2 ) −
3
3
Be sure to simplify everything inside of the integral before you integrate. It’s just another power
rule in this case:
Z
√
2
4 2
x ln(x2 ) dx = x3/2 ln(x2 ) − ( x3/2 ) + C
3
3 3
8
2
= x3/2 ln(x2 ) − x3/2 + C
3
9
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2
MA 16021
Integration by parts summary and examples
Z
Example 2. Find
K. Rotz
x sec2 x dx.
Solution: As always, convince yourself that you need to do integration by parts. It’s a product of
functions, so definitely either u-sub or integration by parts is needed. You can rewrite it as
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x(sec x)2 dx.
The second factor does indeed involve a composition – sec x composed with a square – so you might
think u-sub is appropriate. However, using u-sub you’d want to pick u = sec x (the stuff inside the outer
function), in which case du = sec x tan x dx. That du definitely doesn’t appear as a multiplicative factor
of the integral, so u-sub probably isn’t the best way to go.
1. The factors are x (Algebraic) and sec2 x (Trigonometric). By LATE, choose
dv = sec2 x dx.
u = x,
2. Differentiate u and integrate dv:
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du = 1 dx,
v=
sec2 x dx = tan x.
3. From the IBP formula,
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v du
uv
z }| { Z z }| {
2
x sec x dx = x tan x − tan x dx
(1)
In order to integrate tan x, you should ask yourself: do I know of any trig function whose derivative
is tan x? The answer to this question is no, so you should rewrite tan x in terms of sines and cosines:
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Z
sin x
tan x dx =
dx.
cos x
We can do this using a u-sub: if u = cos x (remember: typically you choose the denominator
du
before the numerator in u-sub), then du = − sin x dx, or dx = − sin
. Therefore
x
dx
Z
Z
tan x dx =
z }| {
Z
sin x
du
1
(−
)=−
du.
u
sin x
u
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MA 16021
Integration by parts summary and examples
K. Rotz
We can finally do the integration step since there are no more products or quotients:
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1
−
du = − ln |u| + C = − ln | cos x|.
u
Insert this back into equation (1) above and add on the +C:
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x sec2 x dx = x tan x + ln | cos x| + C .
Note that the second term became positive since we were subtracting off
(1).
4
R
tan x dx in equation