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Course: Calculus 2 > Unit 2

Lesson 1: Integrating with u-substitution

𝘶-substitution

𝘶-Substitution essentially reverses the chain rule for derivatives. In other words, it helps us integrate composite functions.

When finding antiderivatives, we are basically performing "reverse differentiation." Some cases are pretty straightforward. For example, we know the derivative of

x^{2}

2 x

, so

\int 2 x d x = x^{2} + C

. We can use this straightforward reasoning with other basic functions, like

\sin (x)

e^{x}

\frac{1}{x}

, etc.

Other cases, however, are not that simple. For example, what is

\int \cos (3 x + 5) d x

? Hint: it's not

\sin (3 x + 5) + C

. Try differentiating that and you will see why.

One method that can be very useful is $u$ ‍-substitution, which basically reverses the chain rule.

Using $u$ ‍-substitution with indefinite integrals

Imagine we are asked to find

\int 2 x \cos (x^{2}) d x

. Notice that

2 x

is the derivative of

x^{2}

, which is the "inner" function in the composite function

\cos (x^{2})

. In other words, letting

u (x) = x^{2}

and

w (x) = \cos (x)

, we have:

\overset{u^{'}}{\overset{⏞}{2 x}} \underset{w}{\underset{⏟}{\cos (\overset{u}{\overset{⏞}{x^{2}}})}} = u^{'} (x) w (u (x))

This suggests that

u

-substitution is called for. Let's see how it's done.

First, we differentiate the equation

u = x^{2}

according to

x

, while treating

u

as an implicit function of

x

\begin{aligned} u & = x^{2} \\ \frac{d}{d x} [u] & = \frac{d}{d x} [x^{2}] \\ \frac{d u}{d x} & = 2 x \\ d u & = 2 x d x \end{aligned}

In that last row we multiplied the equation by

d x

d u

is isolated. That's somewhat unorthodox, but useful for our next step. So we have

u = x^{2}

and

d u = 2 x d x

. Now we can perform a substitution in the integral:

\begin{aligned} \int 2 x \cos (x^{2}) d x \\ = \int \cos (\underset{u}{\underset{⏟}{x^{2}}}) \underset{d u}{\underset{⏟}{2 x d x}} & Rearrange. \\ = \int \cos (u) d u & Substitute. \end{aligned}

After the substitution we are left with an expression for the antiderivative of

\cos (u)

in terms of

u

. How convenient!

\cos (u)

is a basic function so we can find its antiderivative in a straightforward way. The only thing left to do is return the function to be in terms of

x

\begin{aligned} \int \cos (u) d u \\ = \sin (u) + C \\ = \sin (x^{2}) + C \end{aligned}

In conclusion,

\int 2 x \cos (x^{2}) d x

\sin (x^{2}) + C

. You can differentiate

\sin (x^{2}) + C

to verify that this is true.

Key takeaway #1:

u

-substitution is really all about reversing the chain rule:

According to the chain rule, the derivative of $w (u (x))$ ‍ is $w^{'} (u (x)) \cdot u^{'} (x)$ ‍.
In $u$ ‍-substitution, we take an expression of the form $w^{'} (u (x)) \cdot u^{'} (x)$ ‍ and find its antiderivative $w (u (x))$ ‍.

Key takeaway #2:

u

-substitution helps us take a messy expression and simplify it by making the "inner" function the variable.

Problem 1.A

Problem set 1 will walk you through all the steps of finding the following integral using

u

-substitution.

\int (6 x^{2}) (2 x^{3} + 5)^{6} d x = ?

How should we define $u$ ‍?

Common mistake: getting incorrect expressions for $u$ ‍ or $d u$ ‍

Choosing the wrong expression for

u

will result in a wrong answer. For example, in Problem set 1,

u

must be defined as

2 x^{3} + 5

. Letting

u

6 x^{2}

(2 x^{3} + 5)^{6}

will never work.

Remember: For

u

-substitution to apply, we must be able to write the integrand as

w (u (x)) \cdot u^{'} (x)

. Then,

u

must be defined as the inner function of the composite factor.

Another crucial step in this process is finding

d u

. Make sure you are differentiating

u

correctly, because a wrong expression for

d u

will also result in a wrong answer.

Problem 2

Tim was asked to find

\int \cos (5 x - 7) d x

. This is his work:

\int \cos (5 x - 7) d x = \sin (5 x - 7) + C

Is Tim's work correct? If not, what is his mistake?

(Choice A)
Tim's work is correct.
(Choice B)
Tim's work is incorrect. The antiderivative of $\cos (x)$ ‍ isn't $\sin (x)$ ‍.
(Choice C)
Tim's work is incorrect. He ignored the fact that $\cos (5 x - 7)$ ‍ is a composite function.
(Choice D)
Tim's work is incorrect. You shouldn't add a constant $C$ ‍ when finding an indefinite integral.

Common mistake: not realizing $u$ ‍-substitution is called for

Remember: When integrating a composite function, we can't simply take the antiderivative of the outer function. We need to use

u

-substitution.

Letting

W

be an antiderivative of

w

, this point can be expressed mathematically as follows:

\int w (u (x)) d x \neq W (u (x)) + C

Another common mistake: confusing the inner function and its derivative

Imagine you're trying to find

\int x^{2} \cos (2 x) d x

. You might say "since

2 x

is the derivative of

x^{2}

, we can use

u

-substitution." Actually, since

u

-substitution requires taking the derivative of the inner function,

x^{2}

must be the derivative of

2 x

for

u

-substitution to work. Since that's not the case,

u

-substitution doesn't apply here.

Sometimes we need to multiply/divide the integral by a constant.

Imagine we are asked to find

\int \sin (3 x + 5) d x

. Notice that while we have a composite function

\sin (3 x + 5)

, it is not multiplied by anything. That might seem weird at first, but let's proceed and see what happens.

We let

u = 3 x + 5

, then

d u = 3 d x

. Now we substitute

u

into the integral, not before we perform this clever manipulation:

\int \sin (3 x + 5) d x = \frac{1}{3} \int \sin (3 x + 5) 3 d x

See what we did there? In order to have

3 d x

in the integrand, we multiplied the entire integral by

\frac{1}{3}

. That way we allowed for

u

-substitution while keeping the value of the integral the same.

Let's continue with the substitution:

\begin{aligned} \frac{1}{3} \int \sin (\underset{u}{\underset{⏟}{3 x + 5}}) \underset{d u}{\underset{⏟}{3 d x}} \\ = \frac{1}{3} \int \sin (u) d u \\ = - \frac{1}{3} \cos (u) + C \\ = - \frac{1}{3} \cos (3 x + 5) + C \end{aligned}

Key takeaway: Sometimes we need to multiply or divide the entire integral by a constant, so we can achieve the appropriate form for

u

-substitution without changing the value of the integral.

Problem 3

\int (2 x + 7)^{3} d x = ?

Want more practice? Try this exercise.

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Calculus 2