Download MA 2232 Lecture 12 - More on Integration by Parts Monday

MA 2232 Lecture 12 - More on Integration by Parts
Monday, February 22, 2016.
Objectives: Introduce standard notation for integration by parts, and do more examples.
Homework 12
The Homework is formatted a little different today. Do the numbered problems like we do in the labs.
Our integration by parts formula looks like this.
Z
Z
(1)
u(x)v0 (x) dx = u(x)v(x) − v(x)u0 (x) dx.
In our substitution notation, we used du for u0 (x) dx, and that’s what most people do for integration by
parts. In particular, we’ll let du = u0 (x) dx (just like in substitution), and then
1.
dv = what?.
Substituting these expressions into the integration by parts formula makes the formula much shorter.
Z
Z
(2)
u dv = uv − v du.
Example. For the integral
Z
(3)
x cos(x) dx,
we would write
u=x
du = dx
(4)
dv = cos(x) dx
v = sin(x),
which is similar to what we would do in using substitution. Then we get
Z
Z
Z
(5)
x cos(x) dx = uv − v du = x · sin(x) − sin(x) · dx
(6)
2.
= x sin(x) + cos(x) + C.
Use this new notation on
R
x · ex dx.
Example. Since taking a derivative of a polynomial gives us a polynomial with smaller degree, we can
always do an integral of a polynomial times a sine, cosine, or exponential. For example, consider the integral
Z
(7)
(x2 + 3) sin(5x) dx.
We’d split up the integral with
u = x2 + 3
du = 2x dx
(8)
dv = sin(5x) dx
− cos(5x)
v=
,
5
which gives us
(9)
Z
(x2 + 3) sin(5x) dx = −
(x2 + 3) cos(5x)
+
5
Z
Then we use integration by parts again using
(10)
dv = cos(5x)
dx
5
sin(5x)
v = 25 .
u = 2x
du = 2 dx
1
2x cos(5x)
dx.
5
MA 2232 Lecture 12 - More on Integration by Parts
2
This gives us
(11)
−
(x2 + 3) cos(5x)
+
5
Z
2x cos(5x)
(x2 + 3) cos(5x)
2x sin(5x)
2 sin(5x)
dx = −
+
−
dx
5
5
25
25
(x2 + 3) cos(5x) 2x sin(5x) 2 cos(5x)
+
+
+C
=−
5
25
125
Z
(12)
3.
Use integration by parts twice on
R
x2 · ex dx.
Example. Integrating twice can yield interesting results. Consider the integral
Z
(13)
ex sin(2x) dx.
It doesn’t really matter how we separate the two functions, but I’ll let u = sin(2x).
u = sin(2x)
du = 2 cos(2x) dx
(14)
dv = ex dx
v = ex .
This gives us
Z
(15)
ex sin(2x) dx = ex sin(2x) −
Z
2ex cos(2x) dx.
Then we do basically the same thing again,
u = 2 cos(2x)
du = −4 sin(2x) dx
(16)
dv = ex dx
v = ex ,
which gives us
(17)
e sin(2x) −
x
Z
2e cos(2x) dx = e sin(2x) − 2e cos(2x) +
Z
4e sin(2x) dx
= ex sin(2x) − 2ex cos(2x) − 4
Z
ex sin(2x) dx.
(18)
x
x
x
x
At first glance, this looks like we ended up where we started, but we have the equation
Z
Z
x
x
x
(19)
e sin(2x) dx = e sin(2x) − 2e cos(2x) − 4 ex sin(2x) dx.
We can solve for the integral we want (think about the +C).
Z
(20)
5 ex sin(2x) dx = ex sin(2x) − 2ex cos(2x) + C
Z
ex sin(2x) − 2ex cos(2x)
(21)
ex sin(2x) dx =
+C
5
Differentiating the right side will confirm that this is correct.
R
4.
Use this same trick on ex · cos(x) dx.
Example. Now check this one.
(22)
Z
x5 ln(x) dx.
Your first instinct should be to let u = x5 , but that won’t work very well. Look what happens, when we do
it the other way. We use
(23)
u = ln(x)
du = x1 dx
dv = x5 dx
6
v = x6 ,
MA 2232 Lecture 12 - More on Integration by Parts
3
and then get
Z
(24)
(25)
(26)
Z 6
x6 ln(x)
x
−
dx
6
6x
Z 5
x6 ln(x)
x
=
−
dx
6
6
x6 ln(x) x6
=
−
+C
6
36
x5 ln(x) dx =
For relatively simple problems, like the ones you might see in a calculus class, you want your u to be a
function that simplifies as you take a derivative, and your dv shouldn’t get too much worse as you take an
anti-derivative. There is a thing called the Rule of LIATE, which lays out a general hierarchy for what’s a
good u and a good dv. The best u’s are log functions, and the best dv’s are the exponentials. The order is
logs - inverse trigs/hyperbolics - algebraic - trigs - exponentials
I don’t think this is all that important. You get kind of used to the basic forms anyway, and the really
interesting applications of integration by parts probably don’t follow this rule.
R
5.
If you remember our very first example from yesterday, we did ln(x) dx. Do it again. You’ll let
dv = dx, which seems weird, but it works.
R 3
6.
x ln(x) dx.
7.
R
ln(x)
x2
dx.