Calc 1 Lecture Notes
Section 4.6
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Section 4.6: Integration by Substitution
Big idea: Integration by substitution is a formal way of performing “the chain rule in reverse.”
Big skill: You should be able to calculate integrals using substitution.
Let’s say that f(x) is a function that has an antiderivative F(x), which means
d
 f  x   dx  F  x   c and dx F  x   f  x  .
Now let’s look at what happens when we take the derivative of F composed with another
function u:
du  x 
d
F u  x   f u  x  
 f u  x    u  x 
dx
dx
So we see that F  u  x   is the antiderivative of f  u  x    u  x  , which can be written as
 u  x   f u  x   dx  F u  x    c .
Thus, when we see an integrand that is the product of a composition of functions and the
derivative of the inner function, we’ll know that its antiderivative is simply the antiderivative of
the outer function (still composed with the inner function).
Example:  2 xe x dx 
2
Sometimes it is hard to recognize the inner function of the composition, and so we can formalize
the process of “running the chain rule in reverse” by explicitly substituting the inner function
with a new variable u. The goal in this process is to replace all occurrences of x in the integral
with the variable u.
Integration by Substitution steps:
1. Figure out the “inner” function; call it u(x).
du
2. Compute
.
dx
3. Replace all expressions involving the variable x and dx with the new variable u and du.
du
 dx to replace the differential dx.
Use the differential formula du 
dx
4. Evaluate the resulting “u” integral. If you can’t evaluate the integral, try a different
choice of u.
5. Replace all occurrences of the variable u in the antiderivative with the appropriate
function of x.
Calc 1 Lecture Notes
Section 4.6
Practice:
1.
 x x
2.
 x sin  x  dx 
3.
 cos  x   2sin  x   1
4.
2
3
 5 dx 
80
2

x3
1  x 
4 2
dx 
2
dx 
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Calc 1 Lecture Notes
Section 4.6
Page 3 of 4
5. Trick: sometimes you have to expand the integrand after making your
substitution.  x 1  xdx 
Now we can prove theorem 6.1:
Practice:
6.  cot  x  dx 
 tanh  x  
1
7.

1  x2
3
dx 
f  x
 f  x  dx  ln f  x   c …
Calc 1 Lecture Notes
Section 4.6
Page 4 of 4
Integration by substitution with definite integrals:
Change the limits of integration to values that correspond to u and then plug those directly into
the “u” antiderivative. i.e.,
Practice:
2
8.
 x x
2
3
 5 dx 
80
1
2
9.
 2te
1
t2 / 2
dt 
x b
ub
xa
u a 
 u  x   f u  x  dx   f u  du  F u 
ub
u a