Lecture 3: Integration by Parts (Stewart 7.1)
Rahul Krishna
1
Review of Last Time
Recall what we did yesterday: we ran the Chain Rule backwards. By integrating the Chain Rule, we arrived
at an identity
Z
f 0 (g (x)) g 0 (x) dx = f (g (x)) + C
Which we interpreted as a method–Integration by Substitution. The gist of the method was something
like this: when confronted with an integral where the integrand can be written as the product of two functions,
one of which is (can be made to look like) a derivative, then if conditions are right, (this means if the
other function is actually a function of the anti-derivative of the …rst one–man that’s a confusing sentence.)
we should try to interpret the integral as a reverse chain rule. The best way to see how the method works
is to see it in action, which we did plenty of last time, and will do plenty of today.
You should keep in mind that this sort of method, while it looks like an algorithm of some sort, really
isn’t. In order to use this technique properly, you often need some clever insight.
[The interpretation of the Chain Rule as a reparameterization. One example, done in many ways: look
2
2
at the function f (x) = x2 ; f (2x) = (2x) ; and f x3 = x3 . Explain what the derivative is doing in the
integral formulation of integration by parts. Rewrite formula above as
Z
f 0 (g) dg = f (g) + C
dg
dx is the new scaled area element. Problem yesterday–the example I did was too hard to
where dg = dx
visualize. OOPS.]
2
Running the Product Rule Backwards
Yesterday we ran the Chain Rule backwards. Today, we do the same for the Product Rule. Recall what it
says:
d
(u (x) v (x)) = u0 (x) v (x) + u (x) v 0 (x)
dx
We can move this around to get
u (x) v 0 (x) =
d
(u (x) v (x))
dx
and integrating both sides, we get
Z
u (x) v 0 (x) dx = u (x) v (x)
v (x) u0 (x)
Z
v (x) u0 (x) dx
If we adopt the notation used above of scaled area elements, i.e. writing f 0 (x) dx = df; then this can written
succinctly as
Z
Z
udv = uv
vdu
There is also a form for de…nite integrals, which is exactly what you think it is
Z b
Z b
b
udv = (uv)a
vdu
a
a
1
This is because the above formula gives a way to write down anti-derivatives, and the limits follow from
Version II of the Fundamental Theorem.
Intitially, it might seem like Integration by Parts and Integration by Substitution are techniques that
apply to the same situation: when we are given an integrand which is a product of two functions f (x) g 0 (x),
one of which is obviously a derivative some other function (g 0 (x) is the derivative of g (x), duh!). But
Integration by Substitution only applies when you can write f (x) as a function of g (x) : However, when it
does work, you are done–you can actually evaluate the original integral.
Integration by Parts will always
R apply in this situation. The only problem is what’s on the right hand
side of the equation, namely uv
vdu has an integral in it! So sometimes the integral you get as a result
of applying Integration by Parts is too hard to evaluate, or itself requires an integration by parts (hopefully
in the correct direction–I’ll explain that comment in an example), in which case we would be stuck. Still,
it’s an incredibly powerful technique, and actually underlies a whole sea of very interesting math.
Example 1 Evaluate
Z
x sin xdx
We take u = x; dv = sin xdx Thus v can be taken to be v = cos x and du = dx: We apply Parts to get
Z
Z
x sin xdx = udv
Z
= uv
vdu
Z
= x ( cos x)
( cos x) dx
Z
= x cos x + cos xdx = x cos x + sin x + C
Let’s check it
d
( x cos x + sin x) =
dx
cos x + x sin x + cos x = x sin x
It worked!
Example 2 (Related) Let’s try a worse looking polynomial next to a trigonometric function. We will try to
integrate
Z
x2 + 2x 1 cos xdx
Take u = x2 + 2x
1; dv = cos xdx; v = sin x; du = 2x + 2: So we get
Z
Z
2
2
x + 2x 1 cos xdx = x + 2x 1 sin x
(2x + 2) sin xdx
Crap! How do we deal with this integral on the right? Well, we integrate by parts again: taking, for the
integral below, u = 2x + 2; dv = sin xdx; we have v = cos x and du = 2dx; so
Z
Z
(2x + 2) sin xdx = (2x + 2) ( cos x)
2 ( cos x) dx
Z
= (2x + 2) cos x + 2 cos xdx = (2x + 2) cos x + 2 sin x + C
So we have, in total
x2 + 2x
1 cos xdx = x2 + 2x
2
= x + 2x
1 sin x
[ (2x + 2) cos x + 2 sin x] + C
1 sin x + (2x + 2) cos x
2
2 sin x + C
Let’s check it
d
x2 + 2x 1 sin x + (2x + 2) cos x 2 sin x
dx
= (2x + 2) sin x + x2 + 2x 1 cos x + 2 cos x (2x + 2) sin x
2
= x + 2x
2 cos x
1 cos x
It worked!
Note that in the calculation above, we were careful not to go backwards. What do I mean? Well, let’s
do the Integration by Parts again, but let’s do it incorrectly.
R 2
Example 3 (Don’t do this!!!!) We were calculating
x + 2x 1 cos xdx; and had used Integratation
by Parts once to get
Z
Z
x2 + 2x 1 cos xdx = x2 + 2x 1 sin x
(2x + 2) sin xdx
R
So we need to use Parts on (2x + 2) sin xdx to try and evaluate this. What if we took u = sin x and
dv = (2x + 2) dx; so v = x2 + 2x and du = cos xdx: Then we get
Z
Z
(2x + 2) sin xdx = x2 + 2x sin x
x2 + 2x cos xdx
Wow, this doesn’t help at all! Not only do we still have an awful integral to evaluate, but it’s e¤ ectively the
same as the integral we started with. In fact, in total we have
Z
Z
x2 + 2x sin x
x2 + 2x cos xdx
x2 + 2x 1 cos xdx = x2 + 2x 1 sin x
Z
= sin x +
x2 + 2x cos xdx
while
R 2 this equality is true, it doesn’t help, does it? It’s essentially the equality
x + 2x 1 cos xdx:
R
x2 + 2x
1 cos xdx =
How can we keep track of which direction is forwards? Well, in the example above, we have a polynomial
times a trig function. A trick to remember: always di¤erentiate a polynomial away if you can.
Time for some more fun examples.
3
3.1
Examples
Polynomials Times Sines and Cosines
This class of examples we’ve already essentially dealt with. We were integrating functions of the form
p (x) sin x; p (x) cos x where p (x) is a polynomial in x: Why were these tractable integrals? Well, integrating
sin x gives cos x and integrating cos x gives sin x: Therefore, if we use integration by parts on something
looking like p (x) sin ax or p (x) cos ax; we will get
Z
Z
1
1
cos ax
p0 (x) ( cos ax) dx
p (x) sin axdx = p (x)
a
a
Z
Z
1
1
p (x) cos ax = p (x) sin ax
p0 (x) sin xdx
a
a
Now, the integrals on the right hand side are polynomials times sin ax or cos ax; but have degree one less.
So if you continue until we’re only left with constants times sin ax or cos ax; then we are done! That means
if you come across and integral like this, even if p (x) has some huge degree, at least you know you could
integrate it (better yet, you could program a compute to integrate it) even though it might take an absurd
number of integrations by parts to get the answer.
3
3.2
Polynomials Times Exponentials
This is almost the same as the previous class of examples. Let me do one: you guys will get the general
technique from following
Example 4
Z
x3 ex dx
We take u = x3 ; dv = ex dx; v = ex ; du = 3x2 dx to get
Z
Z
x3 ex dx = x3 ex
3x2 ex dx
Z
= x3 ex
3x2 ex
6xex dx
Z
3 x
2 x
x
=x e
3x e + 6xe
6ex dx
= x3
3x2 + 6x
6 ex + C
The problem set will have you derive a formula for a (slightly) more general version of this.
3.3
Sines and Cosines Times Exponentials
Let’s do an example.
Example 5
Z
ex sin xdx
This looks hopeless! We don’t have a polynomials, and we know integrating or di¤ erentiating ex and sin x
does practically nothing. Still, we soldier on, and compute, for u = ex ; dv = sin xdx; v = cos x; du = ex dx;
that
Z
Z
ex sin xdx = ex cos x + ex cos xdx
Now, for
R
ex cos xdx; we take u = ex ; dv = cos xdx; v = sin x; du = ex dx: Then
Z
Z
ex cos xdx = ex sin x
ex sin xdx
In total, we have the equation
Z
x
e sin xdx =
so
2
Z
x
x
e cos x + e sin x
ex sin xdx = ex (sin x
and thus, adding the constant of integration
Z
1
ex sin xdx = ex (sin x
2
Z
ex sin xdx
cos x)
cos x) + C
What just happened? We used integration by parts to get a recursion relation.
This class of examples can be pretty tricky, but on Monday I’ll give you a way to solve these integrals
that (I think) is easier.
4
3.4
Polynomials in Trigonometric Functions
Example 6
Z
sin2 xdx
We’re going to do this two ways. First, by parts, taking u = sin x; dv = sin xdx; v =
to get
Z
Z
sin2 xdx = sin x cos x + cos2 xdx
Now, using that sin2 x + cos2 x = 1; we have
Z
sin2 xdx =
so
Z
Z
sin x cos x +
sin2 xdx =
cos x; du = cos xdx;
sin2 xdx
1
1
( sin x cos x + x) + C
2
Second, we can use some trig identities. We know cos 2x = cos2 x sin2 x = 1
This gives
Z
Z
1
(1 cos 2x) dx
sin2 xdx =
2
1
1
x
sin 2x + C
=
2
2
2 sin2 x; so sin2 x =
1 cos 2x
:
2
since sin 2x = 2 sin x cos x; this is the same answer.
The general method proceeds in two possible ways. First, use reduction formulas (see the book p. 467
for the formula, applicable when n 2;
Z
Z
1
n 1
n
n 1
sin xdx =
cos x sin
x+
sinn 2 xdx
n
n
which is nothing more than integration by parts) to get your exponents down to something manageable.
Second, you can use trig identities to reduce exponenets. On Monday I will give you a fast and easy way to
do this method. For now, though, just try to use standard trig identities and methods to do problems.
3.5
The Invisible dv
There’s another trick. As always, let me illustrate it with an example.
Example 7
Take u = ln x; dv = dx; v = x; du =
dx
x :
Z
ln xdx
Then
Z
ln xdx = x ln x
= x ln x
Z
dx
x
x+C
x
Wow, that works!
d
(x ln x
dx
x) = ln x + x
5
1
x
1 = ln x
It’s pretty clear why we call it the invisible dv; since the trick consists of taking dv to simply be dx: Let’s
do another example:
Example 8
Take u = tan
1
x; dv = dx; v = x; du =
Z
tan
1
1
1+x2 :
Z
1
tan
xdx
Then
xdx = x tan
1
x
= x tan
1
x
OK, that’s enough integrating for now.
6
Z
x
1 + x2
1
ln 1 + x2 dx
2