Math 111, Notes from Wednesday, November 27th
General Idea: The idea of u-substitution is that we have a derivative that is the result of doing
the chain rule, and we want to be able to find the original function, i.e. the antiderivative. How
does the chain rule work? Well, we figure out what the general structure is, for example, if we’re
raising something to a power, or taking cosine of it, or exponentiating it, etc. The key is that
whenever we do the chain rule, we multiply by the derivative of that “something”. So if we’re
trying to find the antiderivative, the inverse of this proccess, we need to be aware that one of the
functions that appears is actually just the derivative of another function there. We want to pick
our u to be the function that was originally differentiated, and then our du will account for the
derivative of that function. Hopefully the process will be made clear from the following examples.
Z
1. (5x + 2)10 dx
Solution. We want to pick u = 5x + 2 because the basic structure of this integral is of the form
“something” to the 10th power. We know this u will work out because the derivative of 5x + 2 is
5, which is just a constant. So, if u = 5x + 2, du = 5 dx, and so dx = du/5. This turns the integral
into
Z
10
u
Z
2.
1
du
=
5
5
Z
u10 du =
1 u11
(5x + 2)11
+C =
+ C.
5 11
55
3
x2 ex dx
3
Solution. Here we want to pick u = x3 , because we can integrate ex , but we can’t integrate ex , and
the derivative of x3 is 3x2 , which is just a constant multiple of x2 . So if u = x3 , then du = 3x2 dx,
so x2 dx = du/3. This gives
Z
Z
3.
du
1
e
=
3
3
u
Z
1
1 3
eu du = eu + C = ex + C.
3
3
cos4 x sin x dx
Solution. Again, we want to differentiate between the main thing that’s going on and the derivative
that’s floating around. In this case, that’s the fact that we’re raising something to the 4th power.
So we let u = cos x, and so du = − sin x dx. This gives
Z
4
u (−du) = −
Z
u4 du = −
u5
cos5 x
+C =−
+ C.
5
5
Z
4.
p
x 1 − x2 dx
Solution. Now the main thing that’s happening is that we’re taking the square root of something,
so we let u = 1 − x2 , in which case du = −2x dx, so x dx =
Z
Z
5.
x3
√ du
−1
u
=
−2
2
Z
u1/2 du =
du
−2 .
This gives
−1 u3/2
−1
+C =
(1 − x2 )3/2 + C.
2 3/2
3
p
1 − x2 dx
Solution. Here’s where things get interesting. As you can see, picking u = 1 − x2 is the obvious
choice, but then du = −2x dx as before, but we have an x3 dx that we have to account for, not just
an x dx. Remember, the entire goal is to turn the thing inside the integral into only functions of
u, not of x. So here’s the trick. We have that u = 1 − x2 , which means that x2 = 1 − u. Then the
x3 dx becomes part x dx =
Z
Z
6.
−du
2
and part x2 = 1 − u. So here’s how this plays out:
Z
Z
Z
p
p
−1
1/2 −du
2
2
2
x 1 − x dx = x
1 − x (x dx) = (1 − u)u
=
(u1/2 − u3/2 ) du
2
2
!
(1 − x2 )3/2 (1 − x2 )5/2
−1 u3/2 u5/2
−
+C =−
+
+ C.
=
2
3/2
5/2
3
5
3
x
dx
1 + x4
Solution. OK, here we have sort of the opposite situation. The immediate choice for u might be
1 + x4 , but its derivative is 4x3 , and we only have an x dx. Unlike in the previous exercise, we don’t
have enough powers of x, and we can’t use the same trick as in the last problem without making
the integral even more messy (see for yourself if you want to). So, in this situation, we know what
we want du to be some constant multiple of x dx, which means that u should be something like x2 .
So what happens if we do that? Let’s pick u = x2 . Then du = 2x dx, so x dx = du/2 and we get
R 1 du
1
. Hopefully this looks vaguely familiar, because the derivative of arctan x is 1+x
2 . You
1+u2 2
need to always be on the lookout for this, because inverse trig functions have very benign looking
derivatives, and so they’re easy to forget. So, this will actually work, and here’s how it goes:
Z
1 du
1
=
2
1+u 2
2
Z
1
1
1
du = arctan u + C = arctan(x2 ) + C.
2
1+u
2
2
Z
7.
√
x2
dx
1 − x6
Solution. OK, so now we know to be on the lookout for things that look like derivatives of inverse
trig functions. Again we have the same problem as in the last problem, because if we let u = 1 − x6 ,
then du = −6x5 , and we don’t have that many x’s floating around. So now we need to recognize
that the derivative of arcsin x is
√ 1
.
1−x2
So we want to choose u = x3 , because then u2 = x6 , and
du = 3x2 dx, so x2 dx = du/3. Then we have
Z
1
du
1
√
=
2
3
1−u 3
Z
√
1
1
1
du = arcsin u + C = arcsin(x3 ) + C.
2
3
3
1−u