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Mathematical Expression Editor
In this section we learn to reverse the chain rule by making a substitution.
1 U-Substitution
Suppose that is an anti-derivative of , i.e., . Consider the composition, , where is a
differentiable function. What is the derivative of ? According to the chain
rule, From the point of view of integration, this differentiation equation is
equivalent to the anti-differentiation equation: What this equation would look
like if we used familar functions? We will let and . Composing them gives
Furthermore, the derivative of is, , and an anti-derivative of is, . Putting this
information into the anti-differentiation equation yields: This is most easily
understood by differentiaing the right hand side using the chain rule(!): In other
words, this equation is difficult to reproduce is we are just staring at the
left hand side: The technique for systematically finding the anti-derivative,
is to make a u-substitution which is a three step process. In the first
step, the substitution is declared and the integral is re-written in terms of
the variable as Notice that the left hand-side has the differential and the
right hand side has the differential . We will learn how to convert between
differentials. Step two is to compute the indefinite integral. Assuming we know an
antiderivative for , we get: The third and final step is to replace the variable in the
previous step with , to obtain Thus Refer back to this equation after you
have worked through some examples and problems to review the method of
-substitution.
example 1 Compute
We will make a u-substitution for the inside of the composition : We now determine
the relationship between the differentials and . Differentiatiating gives: This
equation can be written in differential form as For the purpose of our impending
computation, we will express this equation in terms of the original differential, : We
can now proceed with step one and convert the original integral into a simpler one, in
the variable, : Notice that ’ cancelled with its reciprocal. Now for step two,
anti-differentiation: And the coup de grace, step three, we return to the variable :
Recapping,
(problem 1a) Compute: Let , then
Don’t forget the ‘dx’ in your answer for ‘du’
Convert to an integral in the variable :
Don’t forget the ‘du’ in your integral
The final answer in terms of is:
(problem 1b) Compute: . Let , then .
Don’t forget the ‘dx’ in your answer for ‘du’.
Convert to an integral in the variable :
Don’t forget the ‘du’ in your integral.
The final answer in terms of is:
example 2 Compute Let . We have,
and the integral can be written as The last integral can be computed as and by
back substituting, we have Thus, using u-substitution we can conclude that
(problem 2a) Compute . Let , then .
Don’t forget the ‘dx’ in your answer for ‘du’.
Convert to an integral in the variable :
Don’t forget the ‘du’ in your integral.
The final answer in terms of is:
(problem 2b) Compute . Let , then .
Don’t forget the ‘dx’ in your answer for ‘du’.
Convert to an integral in the variable :
Don’t forget the ‘du’ in your integral.
The final answer in terms of is:
example 3 Compute Let . We have,
The last integral can be computed as and by back substituting, we have Thus,
using u-substitution we can conclude that
(problem 3a) Compute . Let , then .
Don’t forget the ‘dx’ in your answer for ‘du’.
Convert to an integral in the variable :
Don’t forget the ‘du’ in your integral.
The final answer in terms of is:
(problem 3b) Compute . Let , then .
Don’t forget the ‘dx’ in your answer for ‘du’.
Convert to an integral in the variable :
Don’t forget the ‘du’ in your integral.
The final answer in terms of is:
example 4 Compute Let . We have,
The last integral can be computed as and by back substituting, we have Thus,
using u-substitution we can conclude that
(problem 4) Compute . Let , then .
Don’t forget the ‘dx’ in your answer for ‘du’.
Convert to an integral in the variable :
Don’t forget the ‘du’ in your integral.
example 5 Compute Let . We have,
The last integral can be computed as and by back substituting, we have Thus,
using u-substitution we can conclude that
(problem 5) Compute . Let , then .
Don’t forget the ‘dx’ in your answer for ‘du’.
Convert to an integral in the variable :
Don’t forget the ‘du’ in your integral.
example 6 Compute We let . We have,
and the integral can be written as The last integral can be computed as and by
back substituting, we have Thus, using u-substitution we can conclude that
(problem 6a) Compute . Let , then .
Don’t forget the ‘dx’ in your answer for ‘du’.
Convert to an integral in the variable :
Don’t forget the ‘du’ in your integral.
(problem 6b) Compute . Let , then .
Don’t forget the ‘dx’ in your answer for ‘du’.
Convert to an integral in the variable :
Don’t forget the ‘du’ in your integral.
example 7 Compute Let . Then we have,
and the integral can be written as \begin{align*} \int 3x^2\sec (x^3)\tan (x^3) \ dx &= \int \sec (x^3)\tan (x^3) \cdot 3x^2 \ dx\\ &= \int \sec u\tan u \ du. \end{align*}
The last integral can be computed as and by back substituting, we have Thus, using
u-substitution we can conclude that
(problem 7a)
Let
Compute
(problem 7b)
Let
Compute
example 8 Compute Let . Then,
and the integral can be written as \begin{align*} \int x^3\cos (x^4) \ dx &= \int \cos (x^4) \cdot x^3\ dx \\ &= \int \cos (x^4)\cdot \tfrac 14 \cdot 4x^3\ dx\\ &= \int \tfrac 14\cos u \ du. \end{align*}
The last integral can be computed as and by back substituting, we have Thus, using
u-substitution we can conclude that
(problem 8)
Let
Compute
example 9 Compute Let . Then we have, and the integral can be written as \begin{align*} \int xe^{-x^2} \ dx &= \int e^{-x^2} \cdot x\ dx \\ &= \int e^{-x^2}( -\tfrac 12)\ (-2x)\ dx \\ &= \int -\tfrac 12 e^u \ du. \end{align*}
The last integral can be computed as and by back substituting, we have Thus, using
u-substitution we can conclude that
(problem 9a)
Let
Compute
(problem 9b)
Let
Compute
example 10 Compute Let . Then we have, and the integral can be written as The
last integral can be computed as and by back substituting, we have Thus, using
u-substitution we can conclude that
(problem 10)
Let
Compute
example 11 Compute Let . Then we have, and the integral can be written as The
last integral can be computed as and by back substituting, we have Thus, using
u-substitution we can conclude that
(problem 11)
Let
Compute
example 12 Compute Let . Then we have, and the integral can be written as The
last integral can be computed as and by back substituting, we have Thus, using
u-substitution we can conclude that
(problem 12)
Let
Compute
example 13 Compute Let . Then we have, and the integral can be written as The
last integral can be computed as and by back substituting, we have Thus, using
u-substitution we can conclude that
(problem 13)
Let
Compute
example 14 Compute Let . Then we have, and the integral can be written as The
last integral can be computed as and by back substituting, we have Thus, using
u-substitution we can conclude that
(problem 14a)
Let
Compute
(problem 14b)
Let
Compute
example 15 Compute First rewrite the integral: Now, let ; then and the integral can
be written as \begin{align*} \int \frac{\sin x}{\cos x} \ dx &= \int \frac{1}{\cos x}\ \sin x \ dx \\ &= - \int \frac{1}{\cos x}\ \big (-\sin x\big ) \ dx\\ &=-\int \frac{1}{u} \ du. \end{align*}
The last integral can be computed as and by back substituting, we have Thus, using
u-substitution we can conclude that
(problem 15)
Rewrite:
Let
Compute
example 16 Compute First rewrite the integral: Distributing in the numerator, we
get Now, let , then and the integral can be rewritten as The last integral can be
computed as and by back substituting, we have Thus, using u-substitution we can
conclude that
Here is a detailed, lecture style video on u-substitution: