Difference between revisions of "009B Sample Midterm 3, Problem 5"

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|'''1.''' Recall the trig identity
 
|'''1.''' Recall the trig identity
 
|-
 
|-
−
|&nbsp; &nbsp; <math style="vertical-align: -3px">\tan^2x+1=\sec^2x</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -3px">\tan^2x+1=\sec^2x</math>
 
|-
 
|-
 
|'''2.''' Recall the trig identity
 
|'''2.''' Recall the trig identity
 
|-
 
|-
−
|&nbsp; &nbsp; <math style="vertical-align: -13px">\sin^2(x)=\frac{1-\cos(2x)}{2}</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\sin^2(x)=\frac{1-\cos(2x)}{2}</math>
 
|-
 
|-
 
|'''3.''' How would you integrate <math style="vertical-align: -1px">\tan x~dx?</math>
 
|'''3.''' How would you integrate <math style="vertical-align: -1px">\tan x~dx?</math>
 
|-
 
|-
 
|
 
|
−
&nbsp; &nbsp; You could use <math style="vertical-align: 0px">u</math>-substitution. First, write <math style="vertical-align: -14px">\int \tan x~dx=\int \frac{\sin x}{\cos x}~dx.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp; You could use <math style="vertical-align: 0px">u</math>-substitution.  
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; First, write <math style="vertical-align: -14px">\int \tan x~dx=\int \frac{\sin x}{\cos x}~dx.</math>
 
|-
 
|-
 
|
 
|
−
&nbsp; &nbsp; Now, let <math style="vertical-align: -5px">u=\cos(x).</math> Then, <math style="vertical-align: -5px">du=-\sin(x)dx.</math> Thus,  
+
&nbsp; &nbsp; &nbsp; &nbsp; Now, let <math style="vertical-align: -5px">u=\cos(x).</math> Then, <math style="vertical-align: -5px">du=-\sin(x)dx.</math>  
 +
|-
 +
|&nbsp; &nbsp; &nbsp; &nbsp; Thus,  
 
|-
 
|-
 
|
 
|
−
&nbsp; &nbsp; <math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
\displaystyle{\int \tan x~dx} & = & \displaystyle{\int \frac{-1}{u}~du}\\
 
\displaystyle{\int \tan x~dx} & = & \displaystyle{\int \frac{-1}{u}~du}\\
 
&&\\
 
&&\\
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|-
 
|-
 
|
 
|
−
&nbsp; &nbsp;<math style="vertical-align: -14px">\int \tan^3x~dx=\int \tan^2x\tan x ~dx.</math>  
+
&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -14px">\int \tan^3x~dx=\int \tan^2x\tan x ~dx.</math>  
 
|-
 
|-
 
|Since <math style="vertical-align: -5px">\tan^2x=\sec^2x-1,</math> we have  
 
|Since <math style="vertical-align: -5px">\tan^2x=\sec^2x-1,</math> we have  
 
|-
 
|-
 
|
 
|
−
&nbsp; &nbsp;<math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
\displaystyle{\int \tan^3x~dx} & = & \displaystyle{\int (\sec^2x-1)\tan x ~dx}\\
 
\displaystyle{\int \tan^3x~dx} & = & \displaystyle{\int (\sec^2x-1)\tan x ~dx}\\
 
&&\\
 
&&\\
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|-
 
|-
 
|
 
|
−
Let <math style="vertical-align: -5px">u=\tan(x).</math> Then, <math style="vertical-align: -1px">du=\sec^2x~dx.</math> So, we have
+
Let <math style="vertical-align: -5px">u=\tan(x).</math>  
 +
|-
 +
|Then, <math style="vertical-align: -1px">du=\sec^2x~dx.</math>  
 +
|-
 +
|So, we have
 
|-
 
|-
 
|
 
|
−
&nbsp; &nbsp; <math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
\displaystyle{\int \tan^3x~dx} & = & \displaystyle{\int u~du-\int \tan x~dx}\\
 
\displaystyle{\int \tan^3x~dx} & = & \displaystyle{\int u~du-\int \tan x~dx}\\
 
&&\\
 
&&\\
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|-
 
|-
 
|
 
|
−
&nbsp; &nbsp;<math style="vertical-align: -14px">\int \tan^3x~dx=\frac{\tan^2x}{2}-\int \frac{\sin x}{\cos x}~dx.</math>
+
&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -14px">\int \tan^3x~dx=\frac{\tan^2x}{2}-\int \frac{\sin x}{\cos x}~dx.</math>
 
|-
 
|-
−
|Now, we let <math style="vertical-align: 0px">u=\cos x.</math> Then, <math style="vertical-align: -1px">du=-\sin x~dx.</math>  
+
|Now, we let <math style="vertical-align: 0px">u=\cos x.</math>  
 +
|-
 +
|Then, <math style="vertical-align: -1px">du=-\sin x~dx.</math>  
 
|-
 
|-
 
|Therefore, we get  
 
|Therefore, we get  
 
|-
 
|-
 
|
 
|
−
&nbsp; &nbsp;<math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 
\displaystyle{\int \tan^3x~dx} & = & \displaystyle{\frac{\tan^2x}{2}+\int \frac{1}{u}~dx}\\
 
\displaystyle{\int \tan^3x~dx} & = & \displaystyle{\frac{\tan^2x}{2}+\int \frac{1}{u}~dx}\\
 
&&\\
 
&&\\
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|Solving for <math style="vertical-align: -5px">\sin^2(x),</math> we get  
 
|Solving for <math style="vertical-align: -5px">\sin^2(x),</math> we get  
 
|-
 
|-
−
|&nbsp; &nbsp; <math style="vertical-align: -13px">\sin^2(x)=\frac{1-\cos(2x)}{2}.</math>
+
|&nbsp; &nbsp; &nbsp; &nbsp; <math style="vertical-align: -13px">\sin^2(x)=\frac{1-\cos(2x)}{2}.</math>
 
|-
 
|-
 
|Plugging this identity into our integral, we get  
 
|Plugging this identity into our integral, we get  
 
|-
 
|-
 
|
 
|
−
&nbsp; &nbsp; <math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
\displaystyle{\int_0^\pi \sin^2x~dx} & = & \displaystyle{\int_0^\pi \frac{1-\cos(2x)}{2}~dx}\\
 
\displaystyle{\int_0^\pi \sin^2x~dx} & = & \displaystyle{\int_0^\pi \frac{1-\cos(2x)}{2}~dx}\\
 
&&\\
 
&&\\
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|-
 
|-
 
|
 
|
−
&nbsp; &nbsp; <math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
 
\displaystyle{\int_0^\pi \sin^2x~dx} & = & \displaystyle{\left.\frac{x}{2}\right|_{0}^\pi-\int_0^\pi \frac{\cos(2x)}{2}~dx}\\
 
\displaystyle{\int_0^\pi \sin^2x~dx} & = & \displaystyle{\left.\frac{x}{2}\right|_{0}^\pi-\int_0^\pi \frac{\cos(2x)}{2}~dx}\\
 
&&\\
 
&&\\
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|For the remaining integral, we need to use <math style="vertical-align: 0px">u</math>-substitution.  
 
|For the remaining integral, we need to use <math style="vertical-align: 0px">u</math>-substitution.  
 
|-
 
|-
−
|Let <math style="vertical-align: -1px">u=2x.</math> Then, <math style="vertical-align: -1px">du=2~dx</math> and <math style="vertical-align: -18px">\frac{du}{2}=dx.</math>  
+
|Let <math style="vertical-align: -1px">u=2x.</math>  
 +
|-
 +
|Then, <math style="vertical-align: -1px">du=2~dx</math> and <math style="vertical-align: -18px">\frac{du}{2}=dx.</math>  
 
|-
 
|-
 
|Also, since this is a definite integral and we are using <math style="vertical-align: 0px">u</math>-substitution,  
 
|Also, since this is a definite integral and we are using <math style="vertical-align: 0px">u</math>-substitution,  
Line 147: Line 159:
 
|-
 
|-
 
|
 
|
−
&nbsp; &nbsp;<math>\begin{array}{rcl}
+
&nbsp; &nbsp; &nbsp; &nbsp;<math>\begin{array}{rcl}
 
\displaystyle{\int_0^\pi \sin^2x~dx} & = & \displaystyle{\frac{\pi}{2}-\int_0^{2\pi} \frac{\cos(u)}{4}~du}\\
 
\displaystyle{\int_0^\pi \sin^2x~dx} & = & \displaystyle{\frac{\pi}{2}-\int_0^{2\pi} \frac{\cos(u)}{4}~du}\\
 
&&\\
 
&&\\
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!Final Answer: &nbsp;  
 
!Final Answer: &nbsp;  
 
|-
 
|-
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|'''(a)''' &nbsp; <math>\frac{\tan^2x}{2}+\ln |\cos x|+C</math>
+
|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>\frac{\tan^2x}{2}+\ln |\cos x|+C</math>
 
|-
 
|-
−
|'''(b)''' &nbsp; <math>\frac{\pi}{2}</math>
+
|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>\frac{\pi}{2}</math>
 
|}
 
|}
  
 
[[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]
 
[[009B_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']]

Revision as of 17:44, 7 February 2017

Evaluate the indefinite and definite 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_0^\pi \sin^2x~dx}


Foundations:  
1. Recall the trig 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 \tan^2x+1=\sec^2x}
2. Recall the trig 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)=\frac{1-\cos(2x)}{2}}
3. How would you integrate 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 \tan x~dx?}

        You could 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.

        First, 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 \tan x~dx=\int \frac{\sin x}{\cos x}~dx.}

        Now, 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.}

        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 \begin{array}{rcl} \displaystyle{\int \tan x~dx} & = & \displaystyle{\int \frac{-1}{u}~du}\\ &&\\ & = & \displaystyle{-\ln(u)+C}\\ &&\\ & = & \displaystyle{-\ln|\cos x|+C.}\\ \end{array}}


Solution:

(a)

Step 1:  
We start by writing

        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 \tan^3x~dx=\int \tan^2x\tan x ~dx.}

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 \tan^2x=\sec^2x-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 \tan^3x~dx} & = & \displaystyle{\int (\sec^2x-1)\tan x ~dx}\\ &&\\ & = & \displaystyle{\int \sec^2x\tan x~dx-\int \tan x~dx.}\\ \end{array}}

Step 2:  
Now, we need to 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 for the first integral.

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=\tan(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=\sec^2x~dx.}
So, we have

       

Step 3:  
For the remaining integral, we also need to 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.
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 \tan^3x~dx=\frac{\tan^2x}{2}-\int \frac{\sin x}{\cos x}~dx.}

Now, 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 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.}
Therefore, 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 \begin{array}{rcl} \displaystyle{\int \tan^3x~dx} & = & \displaystyle{\frac{\tan^2x}{2}+\int \frac{1}{u}~dx}\\ &&\\ & = & \displaystyle{\frac{\tan^2x}{2}+\ln |u|+C}\\ &&\\ & = & \displaystyle{\frac{\tan^2x}{2}+\ln |\cos x|+C.} \end{array}}

(b)

Step 1:  
One of the double angle formulas 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 \cos(2x)=1-2\sin^2(x).}
Solving for 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),} 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^2(x)=\frac{1-\cos(2x)}{2}.}
Plugging this identity into our integral, 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 \begin{array}{rcl} \displaystyle{\int_0^\pi \sin^2x~dx} & = & \displaystyle{\int_0^\pi \frac{1-\cos(2x)}{2}~dx}\\ &&\\ & = & \displaystyle{\int_0^\pi \frac{1}{2}~dx-\int_0^\pi \frac{\cos(2x)}{2}~dx.}\\ \end{array}}

Step 2:  
If we integrate the first integral, 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 \begin{array}{rcl} \displaystyle{\int_0^\pi \sin^2x~dx} & = & \displaystyle{\left.\frac{x}{2}\right|_{0}^\pi-\int_0^\pi \frac{\cos(2x)}{2}~dx}\\ &&\\ & = & \displaystyle{\frac{\pi}{2}-\int_0^\pi \frac{\cos(2x)}{2}~dx.}\\ \end{array}}

Step 3:  
For the remaining integral, we need to 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.}
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.}
Also, since this is a definite integral and we are using 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,
we need to change the bounds of integration.
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 u_1=2(0)=0} 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 u_2=2(\pi)=2\pi.}
So, the 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_0^\pi \sin^2x~dx} & = & \displaystyle{\frac{\pi}{2}-\int_0^{2\pi} \frac{\cos(u)}{4}~du}\\ &&\\ & = & \displaystyle{\frac{\pi}{2}-\left.\frac{\sin(u)}{4}\right|_0^{2\pi}}\\ &&\\ & = & \displaystyle{\frac{\pi}{2}-\bigg(\frac{\sin(2\pi)}{4}-\frac{\sin(0)}{4}\bigg)}\\ &&\\ & = & \displaystyle{\frac{\pi}{2}.}\\ \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 \frac{\tan^2x}{2}+\ln |\cos 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 \frac{\pi}{2}}

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