Difference between revisions of "009B Sample Final 1, Problem 4"

From Grad Wiki
Jump to navigation Jump to search
Line 1: Line 1:
 
<span class="exam"> Compute the following integrals.
 
<span class="exam"> Compute the following integrals.
  
−
<span class="exam">(a) <math>\int e^x(x+\sin(e^x))~dx</math>
+
<span class="exam">(a) &nbsp;<math>\int \frac{t^2}{\sqrt{1-t^6}}~dt</math>
  
−
<span class="exam">(b) <math>\int \frac{2x^2+1}{2x^2+x}~dx</math>
+
<span class="exam">(b) &nbsp;<math>\int \frac{2x^2+1}{2x^2+x}~dx</math>
  
−
<span class="exam">(c) <math>\int \sin^3x~dx</math>
+
<span class="exam">(c) &nbsp;<math>\int \sin^3x~dx</math>
  
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 
!Foundations: &nbsp;  
 
!Foundations: &nbsp;  
 
|-
 
|-
−
|Recall:
+
|'''1.''' Through partial fraction decomposition, we can write the fraction
 
|-
 
|-
−
|'''1.''' Integration by parts tells us that <math style="vertical-align: -12px">\int u~dv=uv-\int v~du</math>.
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -18px">\frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}</math>
 
|-
 
|-
−
|'''2.''' Through partial fraction decomposition, we can write the fraction &nbsp;<math style="vertical-align: -18px">\frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}</math> &nbsp;for some constants <math style="vertical-align: -4px">A,B</math>.
+
|&nbsp; &nbsp; &nbsp; &nbsp;for some constants <math style="vertical-align: -4px">A,B</math>.
 
|-
 
|-
−
|'''3.''' We have the Pythagorean identity <math style="vertical-align: -5px">\sin^2(x)=1-\cos^2(x)</math>.
+
|'''2.''' We have the Pythagorean identity  
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -5px">\sin^2(x)=1-\cos^2(x)</math>.
 
|}
 
|}
  
Line 70: Line 72:
 
!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
−
|First, we add and subtract <math style="vertical-align: 0px">x</math> from the numerator.  
+
|First, we add and subtract &nbsp;<math style="vertical-align: 0px">x</math>&nbsp; from the numerator.  
 
|-
 
|-
 
|So, we have
 
|So, we have
Line 89: Line 91:
 
|Now, we need to use partial fraction decomposition for the second integral.
 
|Now, we need to use partial fraction decomposition for the second integral.
 
|-
 
|-
−
|Since <math style="vertical-align: -5px">2x^2+x=x(2x+1)</math>, we let <math>\frac{1-x}{2x^2+x}=\frac{A}{x}+\frac{B}{2x+1}</math>.
+
|Since &nbsp;<math style="vertical-align: -5px">2x^2+x=x(2x+1),</math>&nbsp; we let &nbsp;<math>\frac{1-x}{2x^2+x}=\frac{A}{x}+\frac{B}{2x+1}.</math>
 
|-
 
|-
−
|Multiplying both sides of the last equation by <math style="vertical-align: -5px">x(2x+1)</math>,
+
|Multiplying both sides of the last equation by &nbsp;<math style="vertical-align: -5px">x(2x+1),</math>
 
|-
 
|-
−
|we get <math style="vertical-align: -5px">1-x=A(2x+1)+Bx</math>.
+
|we get &nbsp;<math style="vertical-align: -5px">1-x=A(2x+1)+Bx.</math>
 
|-
 
|-
−
|If we let <math style="vertical-align: 0px">x=0</math>, the last equation becomes <math style="vertical-align: -1px">1=A</math>.
+
|If we let &nbsp;<math style="vertical-align: 0px">x=0,</math> the last equation becomes &nbsp;<math style="vertical-align: -1px">1=A.</math>
 
|-
 
|-
−
|If we let <math style="vertical-align: -14px">x=-\frac{1}{2}</math>, then we get &thinsp;<math style="vertical-align: -13px">\frac{3}{2}=-\frac{1}{2}\,B</math>. Thus, <math style="vertical-align: 0px">B=-3</math>.
+
|If we let &nbsp;<math style="vertical-align: -14px">x=-\frac{1}{2},</math>&nbsp; then we get &nbsp;<math style="vertical-align: -13px">\frac{3}{2}=-\frac{1}{2}\,B.</math>&nbsp; Thus, &nbsp;<math style="vertical-align: 0px">B=-3.</math>
 
|-
 
|-
−
|So, in summation, we have&thinsp; <math>\frac{1-x}{2x^2+x}=\frac{1}{x}+\frac{-3}{2x+1}</math>.
+
|So, in summation, we have &nbsp;<math>\frac{1-x}{2x^2+x}=\frac{1}{x}+\frac{-3}{2x+1}.</math>
 
|}
 
|}
  
Line 108: Line 110:
 
|-
 
|-
 
|
 
|
−
::<math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 
\displaystyle{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{\int ~dx+\int\frac{1}{x}~dx+\int\frac{-3}{2x+1}~dx}\\
 
\displaystyle{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{\int ~dx+\int\frac{1}{x}~dx+\int\frac{-3}{2x+1}~dx}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{x+\ln x+ \int\frac{-3}{2x+1}~dx}.\\
+
& = & \displaystyle{x+\ln x+ \int\frac{-3}{2x+1}~dx.}\\
 
\end{array}</math>
 
\end{array}</math>
 
|}
 
|}
Line 118: Line 120:
 
!Step 4: &nbsp;
 
!Step 4: &nbsp;
 
|-
 
|-
−
|For the final remaining integral, we use <math style="vertical-align: 0px">u</math>-substitution.  
+
|For the final remaining integral, we use &nbsp;<math style="vertical-align: 0px">u</math>-substitution.  
 
|-
 
|-
−
|Let <math style="vertical-align: -2px">u=2x+1</math>. Then, <math style="vertical-align: 0px">du=2\,dx</math> and&thinsp; <math style="vertical-align: -14px">\frac{du}{2}=dx</math>.
+
|Let &nbsp;<math style="vertical-align: -2px">u=2x+1.</math>&nbsp; Then, &nbsp;<math style="vertical-align: 0px">du=2\,dx</math>&nbsp; and &nbsp;<math style="vertical-align: -14px">\frac{du}{2}=dx.</math>
 
|-
 
|-
 
|Thus, our final integral becomes
 
|Thus, our final integral becomes
 
|-
 
|-
 
|
 
|
−
::<math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 
\displaystyle{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{x+\ln x+ \int\frac{-3}{2x+1}~dx}\\
 
\displaystyle{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{x+\ln x+ \int\frac{-3}{2x+1}~dx}\\
 
&&\\
 
&&\\
 
& = & \displaystyle{x+\ln x+\int\frac{-3}{2u}~du}\\
 
& = & \displaystyle{x+\ln x+\int\frac{-3}{2u}~du}\\
 
&&\\
 
&&\\
−
& = & \displaystyle{x+\ln x-\frac{3}{2}\ln u +C}.\\
+
& = & \displaystyle{x+\ln x-\frac{3}{2}\ln u +C.}\\
 
\end{array}</math>
 
\end{array}</math>
 
|-
 
|-
Line 136: Line 138:
 
|-
 
|-
 
|
 
|
−
::<math>\int \frac{2x^2+1}{2x^2+x}~dx\,=\,x+\ln x-\frac{3}{2}\ln (2x+1) +C.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\int \frac{2x^2+1}{2x^2+x}~dx\,=\,x+\ln x-\frac{3}{2}\ln (2x+1) +C.</math>
 
|}
 
|}
  
Line 144: Line 146:
 
!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
|-
 
|-
−
|First, we write <math style="vertical-align: -13px">\int\sin^3x~dx=\int \sin^2 x \sin x~dx</math>.
+
|First, we write &nbsp;<math style="vertical-align: -13px">\int\sin^3x~dx=\int \sin^2 x \sin x~dx</math>.
 
|-
 
|-
−
|Using the identity <math style="vertical-align: -2px">\sin^2x+\cos^2x=1</math>, we get <math style="vertical-align: -1px">\sin^2x=1-\cos^2x</math>.
+
|Using the identity &nbsp;<math style="vertical-align: -2px">\sin^2x+\cos^2x=1,</math>&nbsp; we get &nbsp;<math style="vertical-align: -1px">\sin^2x=1-\cos^2x.</math>  
 
|-
 
|-
 
|If we use this identity, we have
 
|If we use this identity, we have
 
|-
 
|-
−
| &nbsp; &nbsp; <math style="vertical-align: -13px">\int\sin^3x~dx=\int (1-\cos^2x)\sin x~dx</math>.
+
|&nbsp; &nbsp; &nbsp; &nbsp;<math style="vertical-align: -13px">\int\sin^3x~dx=\int (1-\cos^2x)\sin x~dx</math>.
 
|-
 
|-
 
|
 
|
Line 158: Line 160:
 
!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
|-
 
|-
−
|Now, we proceed by <math>u</math>-substitution. Let <math>u=\cos x</math>. Then, <math style="vertical-align: -1px">du=-\sin x dx</math>.
+
|Now, we proceed by &nbsp;<math>u</math>-substitution.  
 +
|-
 +
|Let &nbsp;<math>u=\cos x.</math>&nbsp; Then, &nbsp;<math style="vertical-align: -1px">du=-\sin x dx.</math>  
 
|-
 
|-
 
|So we have
 
|So we have
 
|-
 
|-
 
|
 
|
−
::<math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 
\displaystyle{\int\sin^3x~dx} & = & \displaystyle{\int -(1-u^2)~du}\\
 
\displaystyle{\int\sin^3x~dx} & = & \displaystyle{\int -(1-u^2)~du}\\
 
&&\\
 
&&\\
Line 178: Line 182:
 
!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
−
|'''(a)''' &nbsp;<math>xe^x-e^x-\cos(e^x)+C</math>
+
|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp;<math>xe^x-e^x-\cos(e^x)+C</math>
 
|-
 
|-
−
|'''(b)''' &nbsp;<math style="vertical-align: -14px">x+\ln x-\frac{3}{2}\ln (2x+1) +C</math>
+
|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp;<math style="vertical-align: -14px">x+\ln x-\frac{3}{2}\ln (2x+1) +C</math>
 
|-
 
|-
−
|'''(c)''' &nbsp;<math style="vertical-align: -14px">-\cos x+\frac{\cos^3x}{3}+C</math>
+
|&nbsp; &nbsp; '''(c)''' &nbsp; &nbsp;<math style="vertical-align: -14px">-\cos x+\frac{\cos^3x}{3}+C</math>
 
|}
 
|}
 
[[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Final_1|'''<u>Return to Sample Exam</u>''']]

Revision as of 10:44, 27 February 2017

Compute the following integrals.

(a)  

(b)  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \frac{2x^2+1}{2x^2+x}~dx}

(c)  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \sin^3x~dx}

Foundations:  
1. Through partial fraction decomposition, we can write the fraction
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}}
       for some constants Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A,B} .
2. We have the Pythagorean identity
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin^2(x)=1-\cos^2(x)} .


Solution:

(a)

Step 1:  
We first distribute to get
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int e^x(x+\sin(e^x))~dx\,=\,\int e^xx~dx+\int e^x\sin(e^x)~dx.}
Now, for the first integral on the right hand side of the last equation, we use integration by parts.
Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=x} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle dv=e^xdx} . Then, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=dx} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v=e^x} .
So, we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\int e^x(x+\sin(e^x))~dx} & = & \displaystyle{\bigg(xe^x-\int e^x~dx \bigg)+\int e^x\sin(e^x)~dx}\\ &&\\ & = & \displaystyle{xe^x-e^x+\int e^x\sin(e^x)~dx}.\\ \end{array}}
Step 2:  
Now, for the one remaining integral, we use Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} -substitution.
Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=e^x} . Then, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=e^xdx} .
So, we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\int e^x(x+\sin(e^x))~dx} & = & \displaystyle{xe^x-e^x+\int \sin(u)~du}\\ &&\\ & = & \displaystyle{xe^x-e^x-\cos(u)+C}\\ &&\\ & = & \displaystyle{xe^x-e^x-\cos(e^x)+C}.\\ \end{array}}

(b)

Step 1:  
First, we add and subtract  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x}   from the numerator.
So, we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{\int\frac{2x^2+x-x+1}{2x^2+x}~dx}\\ &&\\ & = & \displaystyle{\int \frac{2x^2+x}{2x^2+x}~dx+\int\frac{1-x}{2x^2+x}~dx}\\ &&\\ & = & \displaystyle{\int ~dx+\int\frac{1-x}{2x^2+x}~dx}.\\ \end{array}}
Step 2:  
Now, we need to use partial fraction decomposition for the second integral.
Since  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2x^2+x=x(2x+1),}   we let  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1-x}{2x^2+x}=\frac{A}{x}+\frac{B}{2x+1}.}
Multiplying both sides of the last equation by  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x(2x+1),}
we get  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1-x=A(2x+1)+Bx.}
If we let  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=0,} the last equation becomes  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1=A.}
If we let  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=-\frac{1}{2},}   then we get  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{3}{2}=-\frac{1}{2}\,B.}   Thus,  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B=-3.}
So, in summation, we have  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1-x}{2x^2+x}=\frac{1}{x}+\frac{-3}{2x+1}.}
Step 3:  
If we plug in the last equation from Step 2 into our final integral in Step 1, we have

       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{\int ~dx+\int\frac{1}{x}~dx+\int\frac{-3}{2x+1}~dx}\\ &&\\ & = & \displaystyle{x+\ln x+ \int\frac{-3}{2x+1}~dx.}\\ \end{array}}

Step 4:  
For the final remaining integral, we use  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} -substitution.
Let  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=2x+1.}   Then,  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=2\,dx}   and  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{du}{2}=dx.}
Thus, our final integral becomes

       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\int \frac{2x^2+1}{2x^2+x}~dx} & = & \displaystyle{x+\ln x+ \int\frac{-3}{2x+1}~dx}\\ &&\\ & = & \displaystyle{x+\ln x+\int\frac{-3}{2u}~du}\\ &&\\ & = & \displaystyle{x+\ln x-\frac{3}{2}\ln u +C.}\\ \end{array}}

Therefore, the final answer is

       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \frac{2x^2+1}{2x^2+x}~dx\,=\,x+\ln x-\frac{3}{2}\ln (2x+1) +C.}

(c)

Step 1:  
First, we write  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\sin^3x~dx=\int \sin^2 x \sin x~dx} .
Using the identity  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin^2x+\cos^2x=1,}   we get  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sin^2x=1-\cos^2x.}
If we use this identity, we have
       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\sin^3x~dx=\int (1-\cos^2x)\sin x~dx} .
Step 2:  
Now, we proceed by  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u} -substitution.
Let  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=\cos x.}   Then,  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=-\sin x dx.}
So we have

       Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\int\sin^3x~dx} & = & \displaystyle{\int -(1-u^2)~du}\\ &&\\ & = & \displaystyle{-u+\frac{u^3}{3}+C}\\ &&\\ & = & \displaystyle{-\cos x+\frac{\cos^3x}{3}+C}.\\ \end{array}}


Final Answer:  
    (a)    Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle xe^x-e^x-\cos(e^x)+C}
    (b)    Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x+\ln x-\frac{3}{2}\ln (2x+1) +C}
    (c)    Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\cos x+\frac{\cos^3x}{3}+C}

Return to Sample Exam

Retrieved from "https://gradwiki.math.ucr.edu/index.php?title=009B_Sample_Final_1,_Problem_4&oldid=2868"