# 1.5 Substitution  (Page 3/6)

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$\int f\left(g\left(x\right)\right){g}^{\prime }\text{(}x\right)dx=F\left(g\left(x\right)\right)+C.$

Then

$\begin{array}{cc}{\int }_{a}^{b}f\left[g\left(x\right)\right]{g}^{\prime }\text{(}x\right)dx\hfill & ={F\left(g\left(x\right)\right)|}_{x=a}^{x=b}\hfill \\ & =F\left(g\left(b\right)\right)-F\left(g\left(a\right)\right)\hfill \\ & ={F\left(u\right)|}_{u=g\left(a\right)}^{u=g\left(b\right)}\hfill \\ \\ \\ & ={\int }_{g\left(a\right)}^{g\left(b\right)}f\left(u\right)du,\hfill \end{array}$

and we have the desired result.

## Using substitution to evaluate a definite integral

Use substitution to evaluate ${\int }_{0}^{1}{x}^{2}{\left(1+2{x}^{3}\right)}^{5}dx.$

Let $u=1+2{x}^{3},$ so $du=6{x}^{2}dx.$ Since the original function includes one factor of x 2 and $du=6{x}^{2}dx,$ multiply both sides of the du equation by $1\text{/}6.$ Then,

$\begin{array}{ccc}du\hfill & =\hfill & 6{x}^{2}dx\hfill \\ \frac{1}{6}du\hfill & =\hfill & {x}^{2}dx.\hfill \end{array}$

To adjust the limits of integration, note that when $x=0,u=1+2\left(0\right)=1,$ and when $x=1,u=1+2\left(1\right)=3.$ Then

${\int }_{0}^{1}{x}^{2}{\left(1+2{x}^{3}\right)}^{5}dx=\frac{1}{6}{\int }_{1}^{3}{u}^{5}du.$

Evaluating this expression, we get

$\begin{array}{}\\ \\ \frac{1}{6}{\int }_{1}^{3}{u}^{5}du\hfill & =\left(\frac{1}{6}\right)\left(\frac{{u}^{6}}{6}\right){|}_{1}^{3}\hfill \\ & =\frac{1}{36}\left[{\left(3\right)}^{6}-{\left(1\right)}^{6}\right]\hfill \\ & =\frac{182}{9}.\hfill \end{array}$

Use substitution to evaluate the definite integral ${\int }_{-1}^{0}y{\left(2{y}^{2}-3\right)}^{5}dy.$

$\frac{91}{3}$

## Using substitution with an exponential function

Use substitution to evaluate ${\int }_{0}^{1}x{e}^{4{x}^{2}+3}dx.$

Let $u=4{x}^{3}+3.$ Then, $du=8xdx.$ To adjust the limits of integration, we note that when $x=0,u=3,$ and when $x=1,u=7.$ So our substitution gives

$\begin{array}{cc}{\int }_{0}^{1}x{e}^{4{x}^{2}+3}dx\hfill & =\frac{1}{8}{\int }_{3}^{7}{e}^{u}du\hfill \\ \\ & =\frac{1}{8}{e}^{u}{|}_{3}^{7}\hfill \\ & =\frac{{e}^{7}-{e}^{3}}{8}\hfill \\ & \approx 134.568.\hfill \end{array}$

Use substitution to evaluate ${\int }_{0}^{1}{x}^{2}\text{cos}\left(\frac{\pi }{2}{x}^{3}\right)dx.$

$\frac{2}{3\pi }\approx 0.2122$

Substitution may be only one of the techniques needed to evaluate a definite integral. All of the properties and rules of integration apply independently, and trigonometric functions may need to be rewritten using a trigonometric identity before we can apply substitution. Also, we have the option of replacing the original expression for u after we find the antiderivative, which means that we do not have to change the limits of integration. These two approaches are shown in [link] .

## Using substitution to evaluate a trigonometric integral

Use substitution to evaluate ${\int }_{0}^{\pi \text{/}2}{\text{cos}}^{2}\theta \phantom{\rule{0.2em}{0ex}}d\theta .$

Let us first use a trigonometric identity to rewrite the integral. The trig identity ${\text{cos}}^{2}\theta =\frac{1+\text{cos}\phantom{\rule{0.1em}{0ex}}2\theta }{2}$ allows us to rewrite the integral as

${\int }_{0}^{\pi \text{/}2}{\text{cos}}^{2}\theta d\theta ={\int }_{0}^{\pi \text{/}2}\frac{1+\text{cos}\phantom{\rule{0.1em}{0ex}}2\theta }{2}d\theta .$

Then,

$\begin{array}{cc}{\int }_{0}^{\pi \text{/}2}\left(\frac{1+\text{cos}\phantom{\rule{0.1em}{0ex}}2\theta }{2}\right)d\theta \hfill & ={\int }_{0}^{\pi \text{/}2}\left(\frac{1}{2}+\frac{1}{2}\text{cos}\phantom{\rule{0.1em}{0ex}}2\theta \right)d\theta \hfill \\ \\ \\ & =\frac{1}{2}{\int }_{0}^{\pi \text{/}2}d\theta +{\int }_{0}^{\pi \text{/}2}\text{cos}\phantom{\rule{0.1em}{0ex}}2\theta d\theta .\hfill \end{array}$

We can evaluate the first integral as it is, but we need to make a substitution to evaluate the second integral. Let $u=2\theta .$ Then, $du=2d\theta ,$ or $\frac{1}{2}du=d\theta .$ Also, when $\theta =0,u=0,$ and when $\theta =\pi \text{/}2,u=\pi .$ Expressing the second integral in terms of u , we have

$\begin{array}{}\\ \\ \frac{1}{2}{\int }_{0}^{\pi \text{/}2}d\theta +\frac{1}{2}{\int }_{0}^{\pi \text{/}2}\text{cos}\phantom{\rule{0.1em}{0ex}}2\theta d\theta \hfill & =\frac{1}{2}{\int }_{0}^{\pi \text{/}2}d\theta +\frac{1}{2}\left(\frac{1}{2}\right){\int }_{0}^{\pi }\text{cos}\phantom{\rule{0.1em}{0ex}}udu\hfill \\ & =\frac{\theta }{2}{|}_{\theta =0}^{\theta =\pi \text{/}2}+\frac{1}{4}\text{sin}\phantom{\rule{0.1em}{0ex}}u{|}_{u=0}^{u=\theta }\hfill \\ & =\left(\frac{\pi }{4}-0\right)+\left(0-0\right)=\frac{\pi }{4}.\hfill \end{array}$

## Key concepts

• Substitution is a technique that simplifies the integration of functions that are the result of a chain-rule derivative. The term ‘substitution’ refers to changing variables or substituting the variable u and du for appropriate expressions in the integrand.
• When using substitution for a definite integral, we also have to change the limits of integration.

## Key equations

• Substitution with Indefinite Integrals
$\int f\left[g\left(x\right)\right]{g}^{\prime }\text{(}x\right)dx=\int f\left(u\right)du=F\left(u\right)+C=F\left(g\left(x\right)\right)+C$
• Substitution with Definite Integrals
${\int }_{a}^{b}f\left(g\left(x\right)\right)g\text{'}\left(x\right)dx={\int }_{g\left(a\right)}^{g\left(b\right)}f\left(u\right)du$

Why is u -substitution referred to as change of variable ?

2. If $f=g\circ h,$ when reversing the chain rule, $\frac{d}{dx}\left(g\circ h\right)\left(x\right)={g}^{\prime }\text{(}h\left(x\right)\right){h}^{\prime }\text{(}x\right),$ should you take $u=g\left(x\right)$ or $u=h\left(x\right)?$

$u=h\left(x\right)$

In the following exercises, verify each identity using differentiation. Then, using the indicated u -substitution, identify f such that the integral takes the form $\int f\left(u\right)du.$

$\int x\sqrt{x+1}dx=\frac{2}{15}{\left(x+1\right)}^{3\text{/}2}\left(3x-2\right)+C;u=x+1$

$\int \frac{{x}^{2}}{\sqrt{x-1}}dx\left(x>1\right)=\frac{2}{15}\sqrt{x-1}\left(3{x}^{2}+4x+8\right)+C;u=x-1$

$f\left(u\right)=\frac{{\left(u+1\right)}^{2}}{\sqrt{u}}$

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