TBIL-Cal Substitution Method (TI1)

Memorize an integration formula for every possible function.
Attempt to rewrite the integral in the form $\int g^{'} (u) u^{'} d x = g (u) + C .$
Keep differentiating functions until you come across the function you want to integrate.

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Fact 5.1.4.

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By the chain rule,

\frac{d}{d x} [g (u) + C] = g^{'} (u) u^{'} .

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There is a dual integration technique reversing this process, known as the substitution method.

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This technique involves choosing an appropriate function

u

in terms of

x

to rewrite the integral as follows:

\int f (x) d x = \dots = \int g^{'} (u) u^{'} d x = g (u) + C .

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Observation 5.1.5.

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Recall that

\frac{d u}{d x} = u^{'},

and so

d u = u^{'} d x .

This allows for the following common notation:

\int f (x) d x = \dots = \int g^{'} (u) d u = g (u) + C .

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Therefore, rather than dealing with equations like

u^{'} = \frac{d u}{d x} = x^{2},

we will prefer to write

d u = x^{2} d x .

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Activity 5.1.6.

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Consider

\int x^{2} e^{x^{3}} d x,

which we conjectured earlier to be

\frac{1}{3} e^{x^{3}} + C .

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Suppose we decided to let

u = x^{3} .

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(a)

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Compute

\frac{d u}{d x} = ?,

and rewrite it as

d u = ? d x .

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(b)

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This

? d x

doesn’t appear in

\int x^{2} e^{x^{3}} d x

exactly, so use algebra to solve for

x^{2} d x

in terms of

d u .

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(c)

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Replace

x^{2} d x

and

x^{3}

with

u d u

terms to rewrite

\int x^{2} e^{x^{3}} d x

\int \frac{1}{3} e^{u} d u .

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(d)

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Solve

\int \frac{1}{3} e^{u} d u

in terms of

u,

then replace

u

with

x^{3}

to confirm our original conjecture.

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Example 5.1.7.

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Here is how one might write out the explanation of how to find

\int x^{2} e^{x^{3}} d x

from start to finish:

\begin{aligned} \int x^{2} e^{x^{3}} d x & Let & u = x^{3} \\ d u = 3 x^{2} d x \\ \frac{1}{3} d u = x^{2} d x \\ \int x^{2} e^{x^{3}} d x & = \int e^{(x^{3})} (x^{2} d x) \\ = \int e^{u} \frac{1}{3} d u \\ = \frac{1}{3} e^{u} + C \\ = \frac{1}{3} e^{x^{3}} + C \end{aligned}

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Activity 5.1.8.

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Which step of the previous example do you think was the most important?

Choosing $u = x^{3} .$
Finding $d u = 3 x^{2} d x$ and $\frac{1}{3} d u = x^{2} d x .$
Substituting $\int x^{2} e^{x^{3}} d x$ with $\int \frac{1}{3} e^{u} d u .$
Integrating $\int \frac{1}{3} e^{u} d u = \frac{1}{3} e^{u} + C .$
Unsubstituting $\frac{1}{3} e^{u} + C$ to get $\frac{1}{3} e^{x^{3}} + C .$

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Activity 5.1.9.

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Below are two correct solutions to the same integral, using two different choices for

u .

Which method would you prefer to use yourself?

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\begin{aligned} \int x \sqrt{4 x + 4} d x & Let u = x + 1 \\ 4 u = 4 x + 4 \\ x = u - 1 \\ d u = d x \\ \int x \sqrt{4 x + 4} d x & = \int (u - 1) \sqrt{4 u} d u \\ = \int (2 u^{3 / 2} - 2 u^{1 / 2}) d u \\ = \frac{4}{5} u^{5 / 2} - \frac{4}{3} u^{3 / 2} + C \\ = \frac{4}{5} (x + 1)^{5 / 2} \\ - \frac{4}{3} (x + 1)^{3 / 2} + C \end{aligned}

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\begin{aligned} \int x \sqrt{4 x + 4} d x & Let u = \sqrt{4 x + 4} \\ u^{2} = 4 x + 4 \\ x = \frac{1}{4} u^{2} - 1 \\ d x = \frac{1}{2} u d u \\ \int x \sqrt{4 x + 4} d x & = \int (\frac{1}{4} u^{2} - 1) (u) (\frac{1}{2} u d u) \\ = \int (\frac{1}{8} u^{4} - \frac{1}{2} u^{2}) d u \\ = \frac{1}{40} u^{5} - \frac{1}{6} u^{3} + C \\ = \frac{1}{40} (4 x + 4)^{5 / 2} \\ - \frac{1}{6} (4 x + 4)^{3 / 2} + C \end{aligned}

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Activity 5.1.10.

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Suppose we wanted to try the substitution method to find

\int e^{x} \cos (e^{x} + 3) d x .

Which of these choices for

u

appears to be most useful?

$u = x,$ so $d u = d x$
$u = e^{x},$ so $d u = e^{x} d x$
$u = e^{x} + 3,$ so $d u = e^{x} d x$
$u = \cos (x),$ so $d u = - \sin (x) d x$
$u = \cos (e^{x} + 3),$ so $d u = e^{x} \sin (e^{x} + 3) d x$

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Activity 5.1.11.

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Complete the following solution using your choice from the previous activity to find

\int e^{x} \cos (e^{x} + 3) d x .

\begin{aligned} \int e^{x} \cos (e^{x} + 3) d x & Let & u = ? \\ d u = ? d x \\ \int e^{x} \cos (e^{x} + 3) d x & = \int ? d u \\ = \dots \\ = \sin (e^{x} + 3) + C \end{aligned}

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Activity 5.1.12.

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Complete the following integration by substitution to find

\int \frac{x^{3}}{x^{4} + 4} d x .

\begin{aligned} \int \frac{x^{3}}{x^{4} + 4} d x & Let & u = ? \\ d u = ? d x \\ ? d u = ? d x \\ \int \frac{x^{3}}{x^{4} + 4} d x & = \int \frac{?}{?} d u \\ = \dots \\ = \frac{1}{4} \ln | x^{4} + 4 | + C \end{aligned}

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Activity 5.1.13.

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Given that

\int \frac{x^{3}}{x^{4} + 4} d x = \frac{1}{4} \ln | x^{4} + 4 | + C,

what is the value of

\int_{0}^{2} \frac{x^{3}}{x^{4} + 4} d x ?

$\frac{8}{20}$
$- \frac{8}{20}$
$\frac{1}{4} \ln (20) - \frac{1}{4} \ln (4)$
$\frac{1}{4} \ln (4) - \frac{1}{4} \ln (20)$

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Activity 5.1.14.

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What’s wrong with the following computation?

\begin{aligned} \int_{0}^{2} \frac{x^{3}}{x^{4} + 4} d x & Let & u = x^{4} + 4 \\ d u = 4 x^{3} d x \\ \frac{1}{4} d u = x^{3} d x \\ \int_{0}^{2} \frac{x^{3}}{x^{4} + 4} d x & = \int_{0}^{2} \frac{1 / 4}{u} d u \\ = {[\frac{1}{4} \ln | u |]}_{0}^{2} \\ = \frac{1}{4} \ln 2 - \frac{1}{4} \ln 0 \end{aligned}

The wrong $u$ substitution was made.
The antiderivative of $\frac{1 / 4}{u}$ was wrong.
The $x$ values $0, 2$ were plugged in for the variable $u .$

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Example 5.1.15.

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Here’s one way to show the computation of this definite integral by tracking

x

values in the bounds.

\begin{aligned} \int_{0}^{2} \frac{x^{3}}{x^{4} + 4} d x & Let & u = x^{4} + 4 \\ d u = 4 x^{3} d x \\ \frac{1}{4} d u = x^{3} d x \\ \int_{x = 0}^{x = 2} \frac{x^{3}}{x^{4} + 4} d x & = \int_{x = 0}^{x = 2} \frac{1 / 4}{u} d u \\ = {[\frac{1}{4} \ln | u |]}_{x = 0}^{x = 2} \\ = {[\frac{1}{4} \ln | x^{4} + 4 |]}_{x = 0}^{x = 2} \\ = \frac{1}{4} \ln (20) - \frac{1}{4} \ln (4) \end{aligned}

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Example 5.1.16.

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Instead of unsubstituting

u

values for

x

values, definite integrals may be computed by also substituting

x

values in the bounds with

u

values. Use this idea to complete the following solution:

\begin{aligned} \int_{1}^{3} x^{2} e^{x^{3}} d x & Let & u = ? \\ d u = 3 x^{2} d x \\ \frac{1}{3} d u = x^{2} d x \\ \int_{1}^{3} x^{2} e^{x^{3}} d x & = \int_{x = 1}^{x = 3} e^{(x^{3})} (x^{2} d x) \\ = \int_{u = ?}^{u = ?} e^{u} \frac{1}{3} d u \\ = {[\frac{1}{3} e^{u}]}_{?}^{?} \\ = ? \end{aligned}

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Example 5.1.17.

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Here is how one might write out the explanation of how to find

\int_{1}^{3} x^{2} e^{x^{3}} d x

from start to finish by leaving bounds in terms of

x

instead:

\begin{aligned} \int_{1}^{3} x^{2} e^{x^{3}} d x & Let & u = x^{3} \\ d u = 3 x^{2} d x \\ \frac{1}{3} d u = x^{2} d x \\ \int_{1}^{3} x^{2} e^{x^{3}} d x & = \int_{x = 1}^{x = 3} e^{(x^{3})} (x^{2} d x) \\ = \int_{x = 1}^{x = 3} e^{u} \frac{1}{3} d u \\ = {[\frac{1}{3} e^{u}]}_{x = 1}^{x = 3} \\ = {[\frac{1}{3} e^{x^{3}}]}_{x = 1}^{x = 3} \\ = \frac{1}{3} e^{3^{3}} - \frac{1}{3} e^{1^{3}} \\ = \frac{1}{3} e^{27} - \frac{1}{3} e \end{aligned}

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Activity 5.1.18.

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Use substitution to show that

\int_{1}^{4} \frac{e^{\sqrt{x}}}{\sqrt{x}} d x = 2 e^{2} - 2 e .

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Activity 5.1.19.

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Use substitution to show that

\int_{0}^{π / 4} \sin (2 θ) d θ = \frac{1}{2} .

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Activity 5.1.20.

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Use substitution to show that

\int u^{5} (u^{3} + 1)^{1 / 3} d u = \frac{1}{7} (u^{3} + 1)^{7 / 3} - \frac{1}{4} (u^{3} + 1)^{4 / 3} + C .

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Activity 5.1.21.

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Consider

\int (3 x - 5)^{2} d x .

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(a)

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Solve this integral using substitution.

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(b)

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Replace

(3 x - 5)^{2}

with

(9 x^{2} - 30 x + 25)

in the original integral, the solve using the reverse power rule.

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(c)

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Which method did you prefer?

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Activity 5.1.22.

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Consider

\int \tan (x) d x .

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(a)

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Replace

\tan (x)

in the integral with a fraction involving sine and cosine.

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(b)

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Use substitution to solve the integral.

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Subsection 5.1.2 Videos

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Figure 106. Video: Evaluate various integrals via the substitution method

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Note: a

1 / 6

was accidentally forgotten in the last example shown in the video above.

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Subsection 5.1.3 Exercises

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Exercises available at https://tbil.org/calculus/preview/exercises/#/bank/TI1/

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