CHAPTER 5. THE DEFINITE INTEGRAL
5.5
122
U -substitution
(a) Find the derivative
of sin(x2 ).
Z
(b) Find the anti-derivative
x cos(x2 ) dx.
Example 1.
Solution.
(a) We use the chain rule:
d
sin(x2 ) = cos(x2 )(x2 )0
dx
= cos(x2 )2x
(b) We guess and check, starting with our result from part (a):
d
? = x cos(x2 )
dx
d
sin(x2 ) = 2x cos(x2 ) No. But close.
dx
d 1
sin(x2 ) = x cos(x2 )X
dx 2
Writing our answer in the other direction we get
Z
1
cos(x2 )x dx = sin(x2 ) + C.
2
Example 2.
(a) What is
Z
x
e dx? What is
Z
eu du?
du
(b) Let u = 3x2 + 2x + 1. What is
?
dx
du
(c) “Solve”
for du.
Z dx
(d) Turn eu du into a new integral by substituting u with 3x2 + 2x + 1 and du
with your
answer to part (c).
Z
3x2 +2x+1
(e) Find (6x + 2)e
dx.
(f) Double check your answer by taking the derivative.
Solution.
Z
Z
x
x
(a)
e dx = e + C, eu du = eu + C.
du
= 6x + 2.
dx
(c) du = (6x + 2) dx
(d)
(b)
u
Z
eu du =
=
Z
Z
z
}|
{
du
3x2 + 2x + 1 z }| {
e
(6x + 2) dx
e
3x2 +2x+1
(6x + 2) dx
CHAPTER 5. THE DEFINITE INTEGRAL
123
(e)
Z
(6x + 2)e
3x2 +2x+1
dx =
Z
eu du
= eu + C
=e
3x2 +2x+1
+ C.
(f)
3x2 +2x+1
d 3x2 +2x+1
e
+C =e
3x2 + 2x + 1
dx
=e
3x2 +2x+1
0
(6x + 2)X
Z
Example 3.
(a) Find the anti-derivative
sin(3x + 7) dx.
Z
(b) Find the anti-derivative
x7 sin(x8 ) dx.
Z
(c) Find the anti-derivative (ex + x) sin(2ex + x2 ) dx.
Solution. In the following solution we (as usual) go to one extreme in showing
many steps. With a little practice, about half of the steps below will be
unnecessary. With a lot of practice, almost all of the steps will be unneccessary.
(a)
Z
u = 3x + 7
du = (3x + 7)0 dx
= 3 dx
1
dx = du
Z3
sin(3x + 7) dx =
sin(3x
+ 7}) |{z}
dx
| {z
u
Z
1
du
3
1
sin(u) du
3
1
= ( cos(u)) + C
3
1
=
cos(3x + 7)
3
=
(b)
u = x8
du = (x8 )0 dx
= 8x7 dx
1
x7 dx = du
8
CHAPTER 5. THE DEFINITE INTEGRAL
Z
7
8
x sin(x ) dx =
Z
Z
124
7
sin(|{z}
x8 ) x
| {zdx}
1
du
8
u
1
sin(u) du
8
Z
1
=
sin(u) du
8
1
=
cos(u) + C
8
1
=
cos(x8 ) + C
8
=
(c)
Z
u = 2ex + x2
du = (2ex + 2x) dx
du = 2(ex + x) dx
1
(ex + x) dx = du
Z2
(ex + x) sin(2ex + x2 ) dx =
x
sin(2e
+ x}2 ) (ex + x) dx
| {z
| {z }
u
Z
1
du
2
1
sin(u) du
2
Z
1
=
sin(u) du
2
1
=
cos(u)
2
2
=
cos(2ex + x2 ) + C
1
=
Example 4.
Z
(b) Find
(a) Find
x
dx
1 + x4
Z
x3
dx
1 + x4
Solution.
(a) To get an idea of what should be u, we start by looking for what might
be du, i.e. by looking for ? dx, where we could have ? equal the deriavtive
of u:
Z
Z
Z
x3
1
1
3
3
dx
=
x
·
dx
=
· x
| {zdx}
4
4
1 + x4
1
+
x
1
+
x
| {z }
good du?
good du?
To decide which option to use, ask if we could make u equal to something
else in the integral, so that it’s derivative is what we’ve highlighted for
du.
CHAPTER 5. THE DEFINITE INTEGRAL
125
The second option is better: to make x3 dx = du we can pick u = x4 .
u = x4
Z
du = 4x3 dx
1
x3 dx = du
Z4
x3
dx =
1 + x4
1
x3 dx
1 + |{z}
x4 |1{z }
u
4
du
(b) Let’s look at the options again
Z
Z
Z
x
1
1
dx = x ·
dx =
· |{z}
x dx
4
4
4
1+x
1
+
x
1
+
x
| {z }
good du?
good du?
This one is trickier. The key is to remember: we only need to find some
part of the rest of the integral to call u so that it’s derivative shows up
in du. So, are you ever going to find something in x to call u, so that its
1
derivative is
dx? Not likely.
1 + x4
1
Can you find something in
to call u, so that its derivative is
1 + x4
x dx? What would u have to be? It would have to be x2 . Is there an x2
1
in
? Yes: x4 = (x2 )2 .
4
1+x
Z
Example 5. Find
Z
u = x2
du = 2x dx
1
x dx = du
2
x
dx
1 + x4
tan(x) dx. (Note: this gives us the anti-derivative of one of
our last basic functions.)
Solution. We start by rewriting the function as a fraction:
Z
sin(x)
dx.
cos(x)
The obvious choices for u (and the resulting du) are
u = sin(x)
du = cos(x) dx
u = cos(x)
du = sin(x) dx
CHAPTER 5. THE DEFINITE INTEGRAL
126
✓
1
sin(x)
The original integral does not have cos(x) dx, it has
dx since
dx =
cos(x)
cos(x)
◆
1
sin(x) ·
dx . Thus, we need to use u = cos(x) and du = sin(x) du.
cos(x)
Now we should look at the original integral and see if we can find how to
translate all the x’s. Here’s how it looks
Z
( 1) du
sin(x)
dx
cos(x)
| {z }
.
u
If you replace the indicated parts shown with u and du, you should get this
Z
Z
sin(x)
1
dx =
du.
cos(x)
u
You should be able to double check that we’ve done our work right: if you
substitute back cos(x) in place of u, and sin(x) dx in place of du, then you
get the original integral. Now we finish up:
Z
1
du = ln |u| + C
u
= ln | cos(x)| + C
Add this to your list of basic anti-derivative facts
Z
tan(x) dx = ln | cos(x)| + C = ln | sec(x)| + C
Example 6. Find
Z
cos(ln(x))
dx
x
Solution. Your intuition should be that u can equal what is inside of cosine.
Z
u = ln(x)
1
du =
dx
x
u
du
Z
z }| {
cos( ln(x) )
dx
= cos(u) du
x
= sin(u) + C = sin(ln(x)) + C
Example 7. Find
Z
7 sin(x) cos(x)
dx
1 + cos2 (x)
Solution. Some guesses you could make in solving this integral
u = sin(x)
u = 1 + cos2 (x)
CHAPTER 5. THE DEFINITE INTEGRAL
u = cos(x)
127
u = sin(x) cos(x)
2
u = cos (x)
u = 2 sin(x) cos(x)
You do not need to instantly see which of these choices is best, but you would
figure it out by looking at du in each case (i.e. the derivative of what u equals).
Then, you would ask yourself, if you replace parts of the original integral with
u
Z and du, would you get an integral that you can finish? In other words, would
f (u) du be an integral that is on our short list of basic anti-derivatives (c.f.
page ??)?
Hopefully you see that u = 1 + cos2 (x) works. Here’s how
u = 1 + cos2 (x)
du = 2 sin(x) cos(x) dx
1
du = sin(x) cos(x) dx
2
Z
1
du
2
7 sin(x) cos(x)
dx
1 + cos2 (x)
=
=
=
Z
7
1
·
du
u
2
7
ln |u|
2
7
ln(1 + cos2 (x)) + C
2