5.5 Substitution (Page 2/6)

Calculus volume 1 Page 2 / 6

Use substitution to find the antiderivative of $\int 3 x^{2} {(x^{3} - 3)}^{2} d x .$

$\int 3 x^{2} {(x^{3} - 3)}^{2} d x = \frac{1}{3} {(x^{3} - 3)}^{3} + C$

Sometimes we need to adjust the constants in our integral if they don’t match up exactly with the expressions we are substituting.

Using substitution with alteration

Use substitution to find the antiderivative of $\int z \sqrt{z^{2} - 5} d z .$

Rewrite the integral as $\int z {(z^{2} - 5)}^{1 / 2} d z .$ Let $u = z^{2} - 5$ and $d u = 2 z d z .$ Now we have a problem because $d u = 2 z d z$ and the original expression has only $z d z .$ We have to alter our expression for du or the integral in u will be twice as large as it should be. If we multiply both sides of the du equation by $\frac{1}{2} .$ we can solve this problem. Thus,

\begin{matrix} u & = z^{2} - 5 \\ d u & = 2 z d z \\ \frac{1}{2} d u & = \frac{1}{2} (2 z) d z = z d z . \end{matrix}

Write the integral in terms of u , but pull the $\frac{1}{2}$ outside the integration symbol:

\int z {(z^{2} - 5)}^{1 / 2} d z = \frac{1}{2} \int u^{1 / 2} d u .

Integrate the expression in u :

\begin{matrix} \frac{1}{2} \int u^{1 / 2} d u & = (\frac{1}{2}) \frac{u^{3 / 2}}{\frac{3}{2}} + C \\ = (\frac{1}{2}) (\frac{2}{3}) u^{3 / 2} + C \\ = \frac{1}{3} u^{3 / 2} + C \\ = \frac{1}{3} {(z^{2} - 5)}^{3 / 2} + C . \end{matrix}

Got questions? Get instant answers now!

Use substitution to find the antiderivative of $\int x^{2} {(x^{3} + 5)}^{9} d x .$

$\frac{{(x^{3} + 5)}^{10}}{30} + C$

Got questions? Get instant answers now!

Using substitution with integrals of trigonometric functions

Use substitution to evaluate the integral $\int \frac{sin t}{{cos}^{3} t} d t .$

We know the derivative of $cos t$ is $− sin t,$ so we set $u = cos t .$ Then $d u = − sin t d t .$ Substituting into the integral, we have

\int \frac{sin t}{{cos}^{3} t} d t = − \int \frac{d u}{u^{3}} .

Evaluating the integral, we get

\begin{matrix} − \int \frac{d u}{u^{3}} & = − \int u^{−3} d u \\ = − (- \frac{1}{2}) u^{−2} + C . \end{matrix}

Putting the answer back in terms of t , we get

\begin{matrix} \int \frac{sin t}{{cos}^{3} t} d t & = \frac{1}{2 u^{2}} + C \\ = \frac{1}{2 {cos}^{2} t} + C . \end{matrix}

Got questions? Get instant answers now!

Use substitution to evaluate the integral $\int \frac{cos t}{{sin}^{2} t} d t .$

$- \frac{1}{sin t} + C$

Got questions? Get instant answers now!

Sometimes we need to manipulate an integral in ways that are more complicated than just multiplying or dividing by a constant. We need to eliminate all the expressions within the integrand that are in terms of the original variable. When we are done, u should be the only variable in the integrand. In some cases, this means solving for the original variable in terms of u . This technique should become clear in the next example.

Finding an antiderivative using u -substitution

Use substitution to find the antiderivative of $\int \frac{x}{\sqrt{x - 1}} d x .$

If we let $u = x - 1,$ then $d u = d x .$ But this does not account for the x in the numerator of the integrand. We need to express x in terms of u . If $u = x - 1,$ then $x = u + 1 .$ Now we can rewrite the integral in terms of u :

\begin{array}{l} \int \frac{x}{\sqrt{x - 1}} d x & = \int \frac{u + 1}{\sqrt{u}} d u \\ = \int \sqrt{u} + \frac{1}{\sqrt{u}} d u \\ = \int (u^{1 / 2} + u^{−1 / 2}) d u . \end{array}

Then we integrate in the usual way, replace u with the original expression, and factor and simplify the result. Thus,

\begin{matrix} \int (u^{1 / 2} + u^{−1 / 2}) d u & = \frac{2}{3} u^{3 / 2} + 2 u^{1 / 2} + C \\ = \frac{2}{3} {(x - 1)}^{3 / 2} + 2 {(x - 1)}^{1 / 2} + C \\ = {(x - 1)}^{1 / 2} [\frac{2}{3} (x - 1) + 2] + C \\ = {(x - 1)}^{1 / 2} (\frac{2}{3} x - \frac{2}{3} + \frac{6}{3}) \\ = {(x - 1)}^{1 / 2} (\frac{2}{3} x + \frac{4}{3}) \\ = \frac{2}{3} {(x - 1)}^{1 / 2} (x + 2) + C . \end{matrix}

Got questions? Get instant answers now!

Use substitution to evaluate the indefinite integral $\int {cos}^{3} t sin t d t .$

$- \frac{{cos}^{4} t}{4} + C$

Got questions? Get instant answers now!

Substitution for definite integrals

Substitution can be used with definite integrals, too. However, using substitution to evaluate a definite integral requires a change to the limits of integration. If we change variables in the integrand, the limits of integration change as well.

Substitution with definite integrals

Let $u = g (x)$ and let $g^{′}$ be continuous over an interval $[a, b],$ and let f be continuous over the range of $u = g (x) .$ Then,

\int_{a}^{b} f (g (x)) g^{'} (x) d x = \int_{g (a)}^{g (b)} f (u) d u .

Although we will not formally prove this theorem, we justify it with some calculations here. From the substitution rule for indefinite integrals, if $F (x)$ is an antiderivative of $f (x),$ we have

<< Chapter < Page Page > Chapter >>

Practice Key Terms 2

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

Source: OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Calculus volume 1' conversation and receive update notifications?

Ask

	7 Physiotherapy Flashcards Set 7 By Rhodes Start Flashcards
	Anthropology Life Cycle By Richley Crapo Start Assignment
	14 Dr Landholt Large Animal Medicine-GI quiz By Brooke Delaney Start Exam
	7 Psychology MCQ 2010 Final Exam By John Gabrieli Start Exam
	3 BOD GI quiz By Brooke Delaney Start Exam
	6 Neuroanatomy 06 Head Somatic Visceral Sensory By Stephen Voron Start Quiz
	2 Psychology MCQ 2009 1 Exam By John Gabrieli Start Exam
	6 Microbiology Midterm Practice By Madison Christian Start Test
	Elementary algebra By OpenStax Read Online Course
	2 Neuroanatomy 02 Brain External Features By Stephen Voron Start Quiz