MathAdmin: Created page with "==Introduction== The method of $u$ -substitution is used to simplify the function you are integrating so that you can easily recog..."

2017-09-25T15:19:30Z

Created page with "==Introduction== The method of <math style="vertical-align: -1px">u</math>-substitution is used to simplify the function you are integrating so that you can easily recog..."

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==Introduction==
The method of  <math style="vertical-align: -1px">u</math>-substitution is used to simplify the function you are integrating so that you can easily recognize it's antiderivative.

This method is closely related to the chain rule for derivatives.

One question that is frequently asked is "How do you know what substitution to make?" In general, this is a difficult question to answer since it is dependent on the integral. The best way to master  <math style="vertical-align: -1px">u</math>-substitution is to work out as many problems as possible. This will help you:

(1) understand the  <math style="vertical-align: -1px">u</math>-substitution method and

(2) correctly identify the necessary substitution.

'''<u>NOTE</u>:''' After you plug-in your substitution, all of the  <math style="vertical-align: 0px">x</math>'s in your integral should be gone. The only variables remaining in your integral should be  <math style="vertical-align: 0px">u</math>'s.

==Warm-Up==
Evaluate the following indefinite integrals.

'''1)'''   <math>\int (8x+5)e^{4x^2+5x+3}~dx</math>

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution:  
|-
|Let  <math style="vertical-align: -3px">u=4x^2+5x+3.</math>  Then,  <math style="vertical-align: -5px">du=(8x+5)~dx.</math>
|-
|-
|Plugging these into our integral, we get  <math style="vertical-align: -14px">\int e^u~du,</math>  which we know how to integrate.
|-
|So, we get
|-
|
::<math>\begin{array}{rcl}
\displaystyle{\int (8x+5)e^{4x^2+5x+3}~dx} & = & \displaystyle{\int e^u~du}\\
&&\\
& = & \displaystyle{e^u+C}\\
&&\\
& = & \displaystyle{e^{4x^2+5x+3}+C.} \\
\end{array}</math>
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       <math>e^{4x^2+5x+3}+C</math>
|-
|}

'''2)'''   <math>\int\frac{x}{\sqrt{1-2x^2}}~dx</math>

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution:  
|-
|Let  <math style="vertical-align: -2px">u=1-2x^2.</math>  Then,  <math style="vertical-align: -2px">du=-4x~dx.</math>  Hence,  <math style="vertical-align: -15px">\frac{du}{-4}=x~dx.</math> 
|-
|Plugging these into our integral, we get
|-
|
::<math>\begin{array}{rcl}
\displaystyle{\int\frac{x}{\sqrt{1-2x^2}}~dx} & = & \displaystyle{\int -\frac{1}{4}~u^{-\frac{1}{2}}~du}\\
&&\\
& = & \displaystyle{-\frac{1}{2}u^{\frac{1}{2}}+C}\\
&&\\
& = & \displaystyle{-\frac{1}{2}\sqrt{1-2x^2}+C.} \\
\end{array}</math>
|-
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       <math>-\frac{1}{2}\sqrt{1-2x^2}+C</math>
|-
|}

'''3)'''   <math>\int\frac{\sin(\ln x)}{x}~dx</math>

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution:  
|-
|Let  <math style="vertical-align: -6px">u=\ln(x).</math>  Then,  <math style="vertical-align: -14px">du=\frac{1}{x}~dx.</math>
|-
|Plugging these into our integral, we get
|-
|
::<math>\begin{array}{rcl}
\displaystyle{\int\frac{\sin(\ln x)}{x}~dx} & = & \displaystyle{\int \sin(u)~du}\\
&&\\
& = & \displaystyle{-\cos(u)+C}\\
&&\\
& = & \displaystyle{-\cos(\ln x)+C.} \\
\end{array}</math>
|-
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       <math>-\cos(\ln x)+C</math>
|-
|}

'''4)'''   <math>\int xe^{x^2}~dx</math>

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Solution:  
|-
|Let  <math style="vertical-align: -1px">u=x^2.</math>  Then,  <math style="vertical-align: -1px">du=2x~dx</math>  and  <math style="vertical-align: -15px">\frac{du}{2}=x~dx.</math> 
|-
|Plugging these into our integral, we get
|-
|
::<math>\begin{array}{rcl}
\displaystyle{\int xe^{x^2}~dx} & = & \displaystyle{\int \frac{1}{2}e^u~du}\\
&&\\
& = & \displaystyle{\frac{1}{2}e^u+C}\\
&&\\
& = & \displaystyle{\frac{1}{2}e^{x^2}+C.} \\
\end{array}</math>
|-
|}

{| class = "mw-collapsible mw-collapsed" style = "text-align:left;"
!Final Answer:  
|-
|       <math>\frac{1}{2}e^{x^2}+C</math>
|-
|}

== Exercise 1 ==

Evaluate the indefinite integral  <math style="vertical-align: -16px">\int \frac{2}{y^2+4}~dy.</math>

First, we factor out  <math style="vertical-align: -1px">4</math>  out of the denominator.

So, we have

::<math>\begin{array}{rcl}
\displaystyle{\int \frac{2}{y^2+4}~dy} & = & \displaystyle{\frac{1}{4}\int \frac{2}{\frac{y^2}{4}+1}~dy}\\
&&\\
& = & \displaystyle{\frac{1}{2}\int \frac{1}{(\frac{y}{2})^2+1}~dy.}\\
\end{array}</math>

Now, we use  <math style="vertical-align: -1px">u</math>-substitution. Let  <math>u=\frac{y}{2}.</math>

Then,  <math style="vertical-align: -14px">du=\frac{1}{2}~dy</math>  and  <math style="vertical-align: -5px">2~du=dy.</math>

Plugging these into our integral, we get

::<math>\begin{array}{rcl}
\displaystyle{\int \frac{2}{y^2+4}~dy} & = & \displaystyle{\frac{1}{2}\int \frac{2}{u^2+1}~du}\\
&&\\
& = & \displaystyle{\int \frac{1}{u^2+1}~du}\\
&&\\
& = & \displaystyle{\arctan(u)+C}\\
&&\\
& = & \displaystyle{\arctan\bigg(\frac{y}{2}\bigg)+C.}\\
\end{array}</math>

So, we have
::<math>\int \frac{2}{y^2+4}~dy=\arctan\bigg(\frac{y}{2}\bigg)+C.</math>

== Exercise 2 ==

Evaluate the indefinite integral  <math style="vertical-align: -17px">\int \frac{\cos(x)}{(5+\sin x)^2}~dx.</math>

Let  <math style="vertical-align: -5px">u=5+\sin(x).</math>  Then,  <math style="vertical-align: -5px">u=\cos(x)~dx.</math>

Plugging these into our integral, we get

::<math>\begin{array}{rcl}
\displaystyle{\int \frac{\cos(x)}{(5+\sin x)^2}~dx} & = & \displaystyle{\int \frac{1}{u^2}~du}\\
&&\\
& = & \displaystyle{-\frac{1}{u}+C}\\
&&\\
& = & \displaystyle{-\frac{1}{5+\sin(x)}+C.}
\end{array}</math>

So, we have
::<math>\int \frac{\cos(x)}{(5+\sin x)^2}~dx=-\frac{1}{5+\sin(x)}+C.</math>

== Exercise 3 ==

Evaluate the indefinite integral  <math style="vertical-align: -16px">\int \frac{x+5}{2x+3}~dx.</math>

Here, the substitution is not obvious.

Let  <math style="vertical-align: -3px">u=2x+3.</math>  Then,  <math style="vertical-align: -1px">du=2~dx</math>  and  <math style="vertical-align: -14px">\frac{du}{2}=dx.</math>

Now, we need a way of getting rid of  <math style="vertical-align: -2px">x+5</math>  in the numerator.

Solving for  <math style="vertical-align: 0px">x</math>  in the first equation, we get  <math style="vertical-align: -14px">x=\frac{1}{2}u-\frac{3}{2}.</math>

Plugging these into our integral, we get

::<math>\begin{array}{rcl}
\displaystyle{\int \frac{x+5}{2x+3}~dx} & = & \displaystyle{\int \frac{(\frac{1}{2}u-\frac{3}{2})+5}{2u}~du}\\
&&\\
& = & \displaystyle{\frac{1}{2}\int \frac{\frac{1}{2}u+\frac{7}{2}}{u}~du}\\
&&\\
& = & \displaystyle{\frac{1}{4}\int \frac{u+7}{u}~du}\\
&&\\
& = & \displaystyle{\frac{1}{4}\int 1+\frac{7}{u}~du}\\
&&\\
& = & \displaystyle{\frac{1}{4}(u+7\ln|u|)+C}\\
&&\\
& = & \displaystyle{\frac{1}{4}(2x+3+7\ln|2x+3|)+C.}\\
\end{array}</math>

So, we get
::<math>\int \frac{x+5}{2x+3}~dx=\frac{1}{4}(2x+3+7\ln|2x+3|)+C.</math>

== Exercise 4 ==

Evaluate the indefinite integral  <math style="vertical-align: -14px">\int \frac{x^2+4}{x+2}~dx.</math>

Let  <math style="vertical-align: -2px">u=x+2.</math>  Then,  <math style="vertical-align: -1px">du=dx.</math>

Now, we need a way of replacing  <math style="vertical-align: -2px">x^2+4.</math>

If we solve for  <math style="vertical-align: 0px">x</math>  in our first equation, we get  <math style="vertical-align: -1px">x=u-2.</math>

Now, we square both sides of this last equation to get  <math style="vertical-align: -5px">x^2=(u-2)^2.</math>

Plugging in to our integral, we get

::<math>\begin{array}{rcl}
\displaystyle{\int \frac{x^2+4}{x+2}~dx} & = & \displaystyle{\int \frac{(u-2)^2+4}{u}~du}\\
&&\\
& = & \displaystyle{\int \frac{u^2-4u+4+4}{u}~du}\\
&&\\
& = & \displaystyle{\int \frac{u^2-4u+8}{u}~du}\\
&&\\
& = & \displaystyle{\int u-4+\frac{8}{u}~du}\\
&&\\
& = & \displaystyle{\frac{u^2}{2}-4u+8\ln|u|+C}\\
&&\\
& = & \displaystyle{\frac{(x+2)^2}{2}-4(x+2)+8\ln|x+2|+C.}\\
\end{array}</math>

So, we have
::<math>\int \frac{x^2+4}{x+2}~dx=\frac{(x+2)^2}{2}-4(x+2)+8\ln|x+2|+C.</math>

U-substitution - Revision history

MathAdmin: Created page with "==Introduction== The method of u-substitution is used to simplify the function you are integrating so that you can easily recog..."

MathAdmin: Created page with "==Introduction== The method of $u$ -substitution is used to simplify the function you are integrating so that you can easily recog..."