U-substitution - Revision history

Kayla Murray at 21:57, 24 August 2017

2017-08-24T21:57:53Z

Kayla Murray at 21:38, 24 August 2017

2017-08-24T21:38:18Z

Kayla Murray at 21:22, 24 August 2017

2017-08-24T21:22:45Z

Kayla Murray at 21:08, 24 August 2017

2017-08-24T21:08:07Z

Kayla Murray at 20:45, 24 August 2017

2017-08-24T20:45:09Z

Kayla Murray at 20:38, 24 August 2017

2017-08-24T20:38:33Z

Kayla Murray at 20:26, 24 August 2017

2017-08-24T20:26:05Z

Kayla Murray at 20:19, 24 August 2017

2017-08-24T20:19:07Z

Kayla Murray: /* Exercise 3 */

2016-06-23T20:52:29Z

Exercise 3

Kayla Murray: /* Exercise 2 */

2016-06-23T20:44:47Z

Exercise 2

@@ Line 210: / Line 210: @@
 == Exercise 4 ==
-Evaluate the indefinite integral &nbsp;<math>\int \frac{x^2+4}{x+2}~dx.</math>
+Evaluate the indefinite integral &nbsp;<math style="vertical-align: -14px">\int \frac{x^2+4}{x+2}~dx.</math>
-Let &nbsp;<math>u=x+2.</math>&nbsp; Then, &nbsp;<math>du=dx.</math>
+Let &nbsp;<math style="vertical-align: -2px">u=x+2.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -1px">du=dx.</math>
-Now, we need a way of replacing &nbsp;<math>x^2+4.</math>
+Now, we need a way of replacing &nbsp;<math style="vertical-align: -2px">x^2+4.</math>
-If we solve for &nbsp;<math>x</math>&nbsp; in our first equation, we get &nbsp;<math>x=u-2.</math>
+If we solve for &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; in our first equation, we get &nbsp;<math style="vertical-align: -1px">x=u-2.</math>
-Now, we square both sides of this last equation to get &nbsp;<math>x^2=(u-2)^2.</math>
+Now, we square both sides of this last equation to get &nbsp;<math style="vertical-align: -5px">x^2=(u-2)^2.</math>
 Plugging in to our integral, we get

@@ Line 179: / Line 179: @@
 == Exercise 3 ==
-Evaluate the indefinite integral &nbsp;<math>\int \frac{x+5}{2x+3}~dx.</math>
+Evaluate the indefinite integral &nbsp;<math style="vertical-align: -16px">\int \frac{x+5}{2x+3}~dx.</math>
 Here, the substitution is not obvious.
-Let &nbsp;<math>u=2x+3.</math>&nbsp; Then, &nbsp;<math>du=2~dx</math>&nbsp; and &nbsp;<math>\frac{du}{2}=dx.</math>
+Let &nbsp;<math style="vertical-align: -3px">u=2x+3.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -1px">du=2~dx</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">\frac{du}{2}=dx.</math>
-Now, we need a way of getting rid of &nbsp;<math>x+5</math>&nbsp; in the numerator.
+Now, we need a way of getting rid of &nbsp;<math style="vertical-align: -2px">x+5</math>&nbsp; in the numerator.
-Solving for &nbsp;<math>x</math>&nbsp; in the first equation, we get &nbsp;<math>x=\frac{1}{2}u-\frac{3}{2}.</math>
+Solving for &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; in the first equation, we get &nbsp;<math style="vertical-align: -14px">x=\frac{1}{2}u-\frac{3}{2}.</math>
 Plugging these into our integral, we get

@@ Line 160: / Line 160: @@
 == Exercise 2 ==
-Evaluate the indefinite integral &nbsp;<math>\int \frac{\cos(x)}{(5+\sin x)^2}~dx.</math>
+Evaluate the indefinite integral &nbsp;<math style="vertical-align: -17px">\int \frac{\cos(x)}{(5+\sin x)^2}~dx.</math>
-Let &nbsp;<math>u=5+\sin(x).</math>&nbsp; Then, &nbsp;<math>u=\cos(x)~dx.</math>
+Let &nbsp;<math style="vertical-align: -5px">u=5+\sin(x).</math>&nbsp; Then, &nbsp;<math style="vertical-align: -5px">u=\cos(x)~dx.</math>
 Plugging these into our integral, we get

@@ Line 127: / Line 127: @@
 == Exercise 1 ==
-Evaluate the indefinite integral &nbsp;<math>\int \frac{2}{y^2+4}~dy.</math>
+Evaluate the indefinite integral &nbsp;<math style="vertical-align: -16px">\int \frac{2}{y^2+4}~dy.</math>
-First, we factor out &nbsp;<math>4</math>&nbsp; out of the denominator.
+First, we factor out &nbsp;<math style="vertical-align: -1px">4</math>&nbsp; out of the denominator.
 So, we have
@@ Line 139: / Line 139: @@
 \end{array}</math>
-Now, we use &nbsp;<math>u</math>-substitution. Let &nbsp;<math>u=\frac{y}{2}.</math>
+Now, we use &nbsp;<math style="vertical-align: -1px">u</math>-substitution. Let &nbsp;<math>u=\frac{y}{2}.</math>
-Then, &nbsp;<math>du=\frac{1}{2}~dy</math>&nbsp; and &nbsp;<math>2~du=dy.</math>
+Then, &nbsp;<math style="vertical-align: -14px">du=\frac{1}{2}~dy</math>&nbsp; and &nbsp;<math style="vertical-align: -5px">2~du=dy.</math>
 Plugging these into our integral, we get

@@ Line 49: / Line 49: @@
 !Solution: &nbsp;
 |-
-|Let &nbsp;<math>u=1-2x^2.</math>&nbsp; Then, &nbsp;<math>du=-4x~dx.</math>&nbsp; Hence, &nbsp;<math>\frac{du}{-4}=x~dx.</math>&nbsp;
+|Let &nbsp;<math style="vertical-align: -2px">u=1-2x^2.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -2px">du=-4x~dx.</math>&nbsp; Hence, &nbsp;<math style="vertical-align: -15px">\frac{du}{-4}=x~dx.</math>&nbsp;
 |-
 |Plugging these into our integral, we get
@@ Line 76: / Line 76: @@
 !Solution: &nbsp;
 |-
-|Let &nbsp;<math>u=\ln(x).</math>&nbsp; Then, &nbsp;<math>du=\frac{1}{x}~dx.</math>
+|Let &nbsp;<math style="vertical-align: -6px">u=\ln(x).</math>&nbsp; Then, &nbsp;<math style="vertical-align: -14px">du=\frac{1}{x}~dx.</math>
 |-
 |Plugging these into our integral, we get
@@ Line 103: / Line 103: @@
 !Solution: &nbsp;
 |-
-|Let &nbsp;<math>u=x^2.</math>&nbsp; Then, &nbsp;<math>du=2x~dx</math>&nbsp; and &nbsp;<math>\frac{du}{2}=x~dx.</math>&nbsp;
+|Let &nbsp;<math style="vertical-align: -1px">u=x^2.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -1px">du=2x~dx</math>&nbsp; and &nbsp;<math style="vertical-align: -15px">\frac{du}{2}=x~dx.</math>&nbsp;
 |-
 |Plugging these into our integral, we get

@@ Line 20: / Line 20: @@
 !Solution: &nbsp;
 |-
-|Let &nbsp;<math>u=4x^2+5x+3.</math>&nbsp; Then, &nbsp;<math>du=(8x+5)~dx</math>.
+|Let &nbsp;<math style="vertical-align: -3px">u=4x^2+5x+3.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -5px">du=(8x+5)~dx.</math>
 |-
 |-
-|Plugging these into our integral, we get &nbsp;<math>\int e^u~du,</math>&nbsp; which we know how to integrate.
+|Plugging these into our integral, we get &nbsp;<math style="vertical-align: -14px">\int e^u~du,</math>&nbsp; which we know how to integrate.
 |-
 |So, we get

@@ Line 1: / Line 1: @@
 ==Introduction==
-The method of <math>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.
+The method of &nbsp;<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.
-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>u</math>-substitution is to work out as many problems as possible. This will help you: (1) understand the <math>u</math>-substitution method and (2) correctly identify the necessary substitution.
+This method is closely related to the chain rule for derivatives.
-<u>NOTE</u>: After you plug-in your substitution, all of the <math>x</math>'s in your integral should be gone. The only variables remaining in your integral should be <math>u</math>'s.
+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 &nbsp;<math style="vertical-align: -1px">u</math>-substitution is to work out as many problems as possible. This will help you:
 ==Warm-Up==
@@ Line 14: / Line 20: @@
 !Solution: &nbsp;
 |-
-|Let <math>u=4x^2+5x+3</math>. Then, <math>du=(8x+5)~dx</math>.
+|Let &nbsp;<math>u=4x^2+5x+3.</math>&nbsp; Then, &nbsp;<math>du=(8x+5)~dx</math>.
 |-
 |-
-|Plugging these into our integral, we get <math>\int e^u~du</math>, which we know how to integrate.
+|Plugging these into our integral, we get &nbsp;<math>\int e^u~du,</math>&nbsp; which we know how to integrate.
 |-
 |So, we get
@@ Line 27: / Line 33: @@
 & = & \displaystyle{e^u+C}\\
 &&\\
-& = & \displaystyle{e^{4x^2+5x+3}+C}. \\
+& = & \displaystyle{e^{4x^2+5x+3}+C.} \\
 \end{array}</math>
 |}
@@ Line 34: / Line 40: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;<math>e^{4x^2+5x+3}+C</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>e^{4x^2+5x+3}+C</math>
 |-
 |}
@@ Line 43: / Line 49: @@
 !Solution: &nbsp;
 |-
-|Let <math>u=1-2x^2</math>. Then, <math>du=-4x~dx</math>. Hence, <math>\frac{du}{-4}=x~dx</math>.
+|Let &nbsp;<math>u=1-2x^2.</math>&nbsp; Then, &nbsp;<math>du=-4x~dx.</math>&nbsp; Hence, &nbsp;<math>\frac{du}{-4}=x~dx.</math>&nbsp;
 |-
 |Plugging these into our integral, we get
@@ Line 53: / Line 59: @@
 & = & \displaystyle{-\frac{1}{2}u^{\frac{1}{2}}+C}\\
 &&\\
-& = & \displaystyle{-\frac{1}{2}\sqrt{1-2x^2}+C}. \\
+& = & \displaystyle{-\frac{1}{2}\sqrt{1-2x^2}+C.} \\
 \end{array}</math>
 |-
@@ Line 61: / Line 67: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;<math>-\frac{1}{2}\sqrt{1-2x^2}+C</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>-\frac{1}{2}\sqrt{1-2x^2}+C</math>
 |-
 |}
@@ Line 70: / Line 76: @@
 !Solution: &nbsp;
 |-
-|Let <math>u=\ln(x)</math>. Then, <math>du=\frac{1}{x}~dx</math>.
+|Let &nbsp;<math>u=\ln(x).</math>&nbsp; Then, &nbsp;<math>du=\frac{1}{x}~dx.</math>
 |-
 |Plugging these into our integral, we get
@@ Line 80: / Line 86: @@
 & = & \displaystyle{-\cos(u)+C}\\
 &&\\
-& = & \displaystyle{-\cos(\ln x)+C}. \\
+& = & \displaystyle{-\cos(\ln x)+C.} \\
 \end{array}</math>
 |-
@@ Line 88: / Line 94: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;<math>-\cos(\ln x)+C</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>-\cos(\ln x)+C</math>
 |-
 |}
@@ Line 97: / Line 103: @@
 !Solution: &nbsp;
 |-
-|Let <math>u=x^2</math>. Then, <math>du=2x~dx</math> and <math>\frac{du}{2}=x~dx</math>.
+|Let &nbsp;<math>u=x^2.</math>&nbsp; Then, &nbsp;<math>du=2x~dx</math>&nbsp; and &nbsp;<math>\frac{du}{2}=x~dx.</math>&nbsp;
 |-
 |Plugging these into our integral, we get
@@ Line 107: / Line 113: @@
 & = & \displaystyle{\frac{1}{2}e^u+C}\\
 &&\\
-& = & \displaystyle{\frac{1}{2}e^{x^2}+C}. \\
+& = & \displaystyle{\frac{1}{2}e^{x^2}+C.} \\
 \end{array}</math>
 |-
@@ Line 115: / Line 121: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp;<math>\frac{1}{2}e^{x^2}+C</math>
+|&nbsp; &nbsp; &nbsp; &nbsp;<math>\frac{1}{2}e^{x^2}+C</math>
 |-
 |}
@@ Line 121: / Line 127: @@
 == Exercise 1 ==
-Evaluate the indefinite integral <math>\int \frac{2}{y^2+4}~dy</math>.
+Evaluate the indefinite integral &nbsp;<math>\int \frac{2}{y^2+4}~dy.</math>
-First, we factor out <math>4</math> out of the denominator.
+First, we factor out &nbsp;<math>4</math>&nbsp; out of the denominator.
 So, we have
@@ Line 130: / Line 136: @@
 \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}.\\
+& = & \displaystyle{\frac{1}{2}\int \frac{1}{(\frac{y}{2})^2+1}~dy.}\\
 \end{array}</math>
-Now, we use <math>u</math>-substitution. Let <math>u=\frac{y}{2}</math>.
+Now, we use &nbsp;<math>u</math>-substitution. Let &nbsp;<math>u=\frac{y}{2}.</math>
-Then, <math>du=\frac{1}{2}~dy</math> and <math>2~du=dy</math>.
+Then, &nbsp;<math>du=\frac{1}{2}~dy</math>&nbsp; and &nbsp;<math>2~du=dy.</math>
 Plugging these into our integral, we get
@@ Line 146: / Line 152: @@
 & = & \displaystyle{\arctan(u)+C}\\
 &&\\
-& = & \displaystyle{\arctan\bigg(\frac{y}{2}\bigg)+C}.\\
+& = & \displaystyle{\arctan\bigg(\frac{y}{2}\bigg)+C.}\\
 \end{array}</math>

← Older revision		Revision as of 20:52, 23 June 2016
Line 174:		Line 174:

	Evaluate the indefinite integral <math>\int \frac{x+5}{2x+3}~dx.</math>		Evaluate the indefinite integral <math>\int \frac{x+5}{2x+3}~dx.</math>
		+
		+	Here, the substitution is not obvious.
		+
		+	Let <math>u=2x+3</math>. Then, <math>du=2~dx</math> and <math>\frac{du}{2}=dx</math>.
		+
		+	Now, we need a way of getting rid of <math>x+5</math> in the numerator.
		+
		+	Solving for <math>x</math> in the first equation, we get <math>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 ==		== Exercise 4 ==

← Older revision		Revision as of 20:44, 23 June 2016
Line 155:		Line 155:

	Evaluate the indefinite integral <math>\int \frac{\cos(x)}{(5+\sin x)^2}~dx.</math>		Evaluate the indefinite integral <math>\int \frac{\cos(x)}{(5+\sin x)^2}~dx.</math>
		+
		+	Let <math>u=5+\sin(x).</math> Then, <math>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 ==		== Exercise 3 ==