Download U-Substitution

U -Substitution
June 23, 2014
U -substitution is a technique used to help integrate a function. Sometimes, it is more necessary than other times. For example, let’s look at these
two integrals:
Z
Z
√
(a) sin(2x + 1) dx
(b) sin x cos x dx
To solve integral (a), we can use u = 2x + 1 and follow
R the process explained
below, but there is an easier
way. We know that sin xdx = − cos x + C,
R
so we can suspect that sin(2x + 1)dx is something like − cos(2x + 1) + C.
However, when we differentiate − cos(2x + 1) + C, we find that (using the
Chain Rule) the derivative is
d
[− cos(2x + 1) + C] = sin(2x + 1) · 2,
(1)
dx
R
not sin(2x + 1). Okay,
so
sin(2x +R 1)dx is not quite − cos(2x + 1) + C.
R
But we do know that kf (x) dx = k f (x) dx if k is a constant and f (x) is
some function, so all we have to do is just divide by that pesky 2 that was
multiplied by sin(2x + 1) in Equation (1):
Z
1
sin(2x + 1) dx = − cos(2x + 1) + C.
2
Indeed, the derivative of − 12 cos(2x + 1) + C is precisely sin(2x + 1) (CHECK
THIS!). You might notice that I did not multiply the C by 12 . That’s because
C is just any constant, and a half of any constant is still just any constant.
Now let’s look at integral (b) above. We don’t√know any elementary
function whose derivative is anything at all like sin x cos x, so we can’t use
the same reasoning. We have to do something else, namely a u-substitution.
The goal of a u-substitution is to simplify the integral and replace it by one
that is more familiar. When we do a u-substitution, we can either write
u = f (x) for some function f (x), write x = g(u) for some function g(u), or
in extreme cases write f (x) = g(u) for some functions f (x) and g(u). In
either case, we want to replace all of the x’s by u’s, including the dx. Here
is how we do this. Let’s just try and set
u = cos x.
1
Then
du
dx
= − sin x. We can also write this as
du = − sin x dx,
(2)
in a sense “multiplying” both sides by dx. In general, in the future whenever
we have u = f (x) or x = g(u), we will write du = f 0 (x) dx or dx = g 0 (u) du,
respectively, so that we can substitute appropriately into the integral.
Now we can also write Equation (2) as
−du = sin x dx.
We want to do this because we notice that the integral we were trying to
solve,
Z
Z
√
√
sin x cos x dx =
cos x sin x dx
has a sin x dx in it, so that we can now substitute into that integral and get
Z
√
u(−du).
This is a much easier integral! Solving, we have
Z
Z
√
√
u(−du) = −
u du
u3/2
+C
3/2
2
= − u3/2 + C
3
2
= − (cos x)3/2 + C
3
=−
The last step is very important: we substituted the cos x back in for u because, after all, we were trying to solve an integral that involved x, not u.
Lastly, we check that indeed
√
d
2
3/2
− (cos x) + C = sin x cos x
dx
3
Hopefully, everything about u-substitution except how I chose the u to
be cos x is clear (if not, please read the process again and, if you still don’t
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understand, make sure you ask me to walk you through it or clear up whatever
is troubling you).
The hard part about u-substitution is in fact the choice of this function
u = f (x), and even knowing whether to write u = f (x) or x = g(u). Two
basic guidelines are the following:
Guideline. (Guidelines for U-Substitution)
(1) If there is a function f (x) whose derivative f 0 (x) is multiplied directly by
dx, try setting u = f (x). If there is more than one such function, try them all.
(2) If a composition g(f (x)) appears in the integrand, try setting u = f (x).
If f (x) satisfies both (1) and (2), then u = f (x) is probably the right choice!
Let’s look at some examples; you should work all of these out so you get
a feeling for how u-substitution works.
R
1. For (sin3 x + 4 sin x) cos x dx, set u = sin x.
R
2. For sin x cos x dx, you can use either u = sin x or u = cos x.
R
3
3. For (lnxx) dx, set u = ln x (note that du = x1 dx).
R x
4. For e2xe +1 dx, set u = ex (note that e2x = u2 and recall the derivative
of tan−1 (x)).
R
5. For (3x + 1)10 dx, you can expand out (3x + 1)10 into a tenth-degree
polynomial by multiplying polynomials together many times (which
takes pretty much forever) or by using Pascal’s triangle (ask me if you
don’t know how), and then integrate each term individually, or you can
simply use the u-substitution u = 3x + 1. I vastly prefer the latter.
Besides, what if the 10 was replaced by 1000?
Now let’s do a harder example:
Z
x
dx
(x + 1)2
We see that there are two functions, x and 1/(x + 1)2 , which are multiplied
by dx. Substituting u = x2 /2 (an antiderivative of x) does not help (try it!).
Let us try substituting an antiderivative of 1/(x + 1)2 , namely
u = −(x + 1)−1
3
(you can find this antiderivative by using the same reasoning as for integral
(a) above; instead of 2x + 1 you now have x + 1) so that
du =
1
dx
(x + 1)2
We have to do some algebraic manipulation to solve for the x in the numerator
of the integrand:
1
u=−
x+1
1
x+1=−
u
x = −1 − u−1
Finally we can substitute to get
Z
(−1 − u−1 )du = −u − ln |u| + C
1
1
+C
− ln −
=
x+1
x + 1
1 1
+C
=
− ln x+1
x + 1
1
=
+ ln |x + 1| + C
x+1
(remember that ln(1/x) = − ln x). Did we get the right answer? Let’s check:
d
1
1
d
+ ln |x + 1| + C = −(x + 1)−2 +
· (x + 1)
dx x + 1
x + 1 dx
−1
1
=
+
(x + 1)2 x + 1
−1
x+1
+
=
2
(x + 1)
(x + 1)2
x
=
.
(x + 1)2
We did! Wow! That took a lot of work and we got a really weird answer,
but we know it’s right. Some problems will be like that.
I feel that there are two types of good math problems: those that are
easy and those that are hard. Those that are easy are meant to give you
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practice and confidence, both of which are really important. Those that are
hard are meant to challenge you, excite you, make you work hard, and feel
really good when you solve them.
Let’s do one more:
Z
1
√
dx
2+x+2 x+1
It is not obvious what substitution to use here. Let’s use
√
u = x + 1.
Then
u2 = x + 1
x = u2 − 1.
(You might also arrive at this substitution
by doing two different substitu√
tions: first v = x + 1 and then u = v.) The point here is that the goal
is to get rid of the square root. When we differentiate both sides (of either
equation above), we get
dx = 2u du
so when we substitute, the integral becomes
Z
Z
u
1
2u du = 2
du
2
2 + u − 1 + 2u
1 + 2u + u2
Z
u
=2
du
(u + 1)2
1
+ ln |u + 1| + C
(from above)
=2
u+1
√
2
=√
+ 2 ln | x + 1 + 1| + C.
x+1+1
So if you thought the previous integral was hard, this one is even harder!
Notice I broke this problem up into two parts, each of which was manageable,
by first solving the previous integral. That’s always a good strategy: if you
can break a problem up into simpler parts, do it! For example, let’s look at
this integral:
Z
1 + 2x
dx
x2 + 1
5
The easiest way to solve this integral is to break it in two:
Z
Z
Z
1 + 2x
1
2x
dx =
dx +
dx
2
2
2
x +1
x +1
x +1
R
R
Now x21+1 dx is just tan−1 x + C, and x22x+1 dx can be solved by using the
substitution u = x2 + 1. Now finish the integral by yourself.
There’s only one more thing left to understand about using a u-substitution,
and that is how to deal with definite integrals. For example,
Z 4
(2x − 4)1006 dx
2
can be solved by using the substitution
u = 2x − 4.
Then
1
du = dx
2
Now what happens to the limits when we substitute back into the integral?
Well, there are two ways to go.
du = 2dx, or
Approach 1. Keep the limits in terms of x.
Z x=4
Z 4
1
1006
1006
u
(2x − 4) dx =
du
2
x=2
2
Z
1 x=4 1006
=
u
du
2 x=2
x=4
1 u1007 =
2 1007 x=2
4
(2x − 4)1007 =
2014
2
1007
(2 · 4 − 4)
(2 · 2 − 4)1007
−
2014
2014
41007
22014
=
=
2014
2014
=
In this approach, we just use u-substitution to find an antiderivative in terms
of x and then use the fundamental theorem of calculus to find the value of
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the definite integral.
Approach 2. Change the limits into limits in terms of u.
We first plug the limits into the equation for u in terms of x:
u = 2(2) − 4 = 0
u = 2(4) − 4 = 4
and
and then we substitute everything:
Z
Z 4
1006
(2x − 4) dx =
2
4
1006
u
0
1
=
2
Z
1
du
2
4
u1006 du
0
1007 4
u
2014 0
41007 01007
=
−
2014 2014
41007
22014
=
=
2014
2014
=
It is important to note that, with the first approach, you should write “x =
2” and “x = 4” while the integrand is in terms of u. This prevents you from
accidentally plugging the numbers into the wrong variable. Why can you not
simply plug 2 and 4 into the antiderivative u1007 /2014? Because x1007 /2014 is
not an antiderivative of (2x−4)1006 , so this violates the fundamental theorem
of calculus! Remember, 2 and 4 are the original limits, so they can only be
plugged into an antiderivative of (2x − 4)1006 .
Okay, fine, but then, why can we plug in 0 and 4 into the antiderivative
u1007 /2014? To not dwell on the technical proof of this statement, let me
just say that it is because, when you do evaluate your integral using the first
approach (which is certainly correct), you end up going through the same
steps as you would in the second approach. Just see for yourself. Didn’t we
have to evaluate 2 · 2 − 4 and 2 · 4 − 4 in both approaches? And didn’t we
have to then plug the numbers into the exact same places?
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