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

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::You could use <math>u</math>-substitution. First, write <math>\int \tan x~dx=\int \frac{\sin x}{\cos x}~dx</math>.
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::You could use <math>u</math>-substitution. First, write <math>\int \tan x~dx=\int \frac{\sin x}{\cos x}~dx.</math>
 
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::Now, let <math>u=\cos(x)</math>. Then, <math>du=-\sin(x)dx</math>.
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::Now, let <math>u=\cos(x)</math>. Then, <math>du=-\sin(x)dx.</math>
 
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::Thus, <math>\int \tan x~dx=\int \frac{-1}{u}~du=-\ln(u)+C=-\ln|\cos x|+C</math>.
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::Thus, <math>\int \tan x~dx=\int \frac{-1}{u}~du=-\ln(u)+C=-\ln|\cos x|+C.</math>
 
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
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|We start by writing <math>\int \tan^3x~dx=\int \tan^2x\tan x ~dx</math>.
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|We start by writing <math>\int \tan^3x~dx=\int \tan^2x\tan x ~dx.</math>  
 
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|Since <math>\tan^2x=\sec^2x-1</math>, we have  
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|Since <math>\tan^2x=\sec^2x-1,</math> we have  
 
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::<math>\int \tan^3x~dx=\int (\sec^2x-1)\tan x ~dx=\int \sec^2x\tan x~dx-\int \tan x~dx</math>.
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::<math>\int \tan^3x~dx=\int (\sec^2x-1)\tan x ~dx=\int \sec^2x\tan x~dx-\int \tan x~dx.</math>  
 
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!Step 2: &nbsp;
 
!Step 2: &nbsp;
 
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|Now, we need to use <math>u</math>-substitution for the first integral. Let <math>u=\tan(x)</math>. Then, <math>du=\sec^2xdx</math>. So, we have
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|Now, we need to use <math>u</math>-substitution for the first integral. Let <math>u=\tan(x).</math> Then, <math>du=\sec^2x~dx.</math> So, we have
 
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::<math>\int \tan^3x~dx=\int u~du-\int \tan x~dx=\frac{u^2}{2}-\int \tan x~dx=\frac{\tan^2x}{2}-\int \tan x~dx</math>.
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::<math>\int \tan^3x~dx=\int u~du-\int \tan x~dx=\frac{u^2}{2}-\int \tan x~dx=\frac{\tan^2x}{2}-\int \tan x~dx.</math>
 
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!Step 3: &nbsp;
 
!Step 3: &nbsp;
 
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|For the remaining integral, we also need to use <math>u</math>-substitution. First, we write <math>\int \tan^3x~dx=\frac{\tan^2x}{2}-\int \frac{\sin x}{\cos x}~dx</math>.
+
|For the remaining integral, we also need to use <math>u</math>-substitution. First, we write <math>\int \tan^3x~dx=\frac{\tan^2x}{2}-\int \frac{\sin x}{\cos x}~dx.</math>
 
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|Now, we let <math>u=\cos x</math>. Then, <math>du=-\sin xdx</math>. So, we get  
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|Now, we let <math>u=\cos x.</math> Then, <math>du=-\sin x~dx.</math> So, we get  
 
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::<math>\int \tan^3x~dx=\frac{\tan^2x}{2}+\int \frac{1}{u}~dx=\frac{\tan^2x}{2}+\ln |u|+C=\frac{\tan^2x}{2}+\ln |\cos x|+C</math>.
+
::<math>\int \tan^3x~dx=\frac{\tan^2x}{2}+\int \frac{1}{u}~dx=\frac{\tan^2x}{2}+\ln |u|+C=\frac{\tan^2x}{2}+\ln |\cos x|+C.</math>
 
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'''(b)'''
 
'''(b)'''
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!Step 1: &nbsp;  
 
!Step 1: &nbsp;  
 
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|One of the double angle formulas is <math>\cos(2x)=1-2\sin^2(x)</math>. Solving for <math>\sin^2(x)</math>, we get <math>\sin^2(x)=\frac{1-\cos(2x)}{2}</math>.
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|One of the double angle formulas is <math>\cos(2x)=1-2\sin^2(x).</math> Solving for <math>\sin^2(x)</math>, we get <math>\sin^2(x)=\frac{1-\cos(2x)}{2}.</math>
 
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|Plugging this identity into our integral, we get  
 
|Plugging this identity into our integral, we get  
 
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::<math>\int_0^\pi \sin^2x~dx=\int_0^\pi \frac{1-\cos(2x)}{2}~dx=\int_0^\pi \frac{1}{2}~dx-\int_0^\pi \frac{\cos(2x)}{2}~dx</math>.
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::<math>\int_0^\pi \sin^2x~dx=\int_0^\pi \frac{1-\cos(2x)}{2}~dx=\int_0^\pi \frac{1}{2}~dx-\int_0^\pi \frac{\cos(2x)}{2}~dx.</math>  
 
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::<math>\int_0^\pi \sin^2x~dx=\left.\frac{x}{2}\right|_{0}^\pi-\int_0^\pi \frac{\cos(2x)}{2}~dx=\frac{\pi}{2}-\int_0^\pi \frac{\cos(2x)}{2}~dx</math>.
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::<math>\int_0^\pi \sin^2x~dx=\left.\frac{x}{2}\right|_{0}^\pi-\int_0^\pi \frac{\cos(2x)}{2}~dx=\frac{\pi}{2}-\int_0^\pi \frac{\cos(2x)}{2}~dx.</math>
 
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!Step 3: &nbsp;
 
!Step 3: &nbsp;
 
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|For the remaining integral, we need to use <math>u</math>-substitution. Let <math>u=2x</math>. Then, <math>du=2dx</math> and <math>\frac{du}{2}=dx</math>. Also, since this is a definite integral  
+
|For the remaining integral, we need to use <math>u</math>-substitution. Let <math>u=2x.</math> Then, <math>du=2~dx</math> and <math>\frac{du}{2}=dx.</math> Also, since this is a definite integral  
 
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|and we are using <math>u</math>-substitution, we need to change the bounds of integration. We have <math>u_1=2(0)=0</math> and <math>u_2=2(\pi)=2\pi</math>.
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|and we are using <math>u</math>-substitution, we need to change the bounds of integration. We have <math>u_1=2(0)=0</math> and <math>u_2=2(\pi)=2\pi.</math>
 
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|So, the integral becomes
 
|So, the integral becomes
 
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::<math>\int_0^\pi \sin^2x~dx=\frac{\pi}{2}-\int_0^{2\pi} \frac{\cos(u)}{4}~du=\frac{\pi}{2}-\left.\frac{\sin(u)}{4}\right|_0^{2\pi}=\frac{\pi}{2}-\bigg(\frac{\sin(2\pi)}{4}-\frac{\sin(0)}{4}\bigg)=\frac{\pi}{2}</math>
+
::<math>\int_0^\pi \sin^2x~dx=\frac{\pi}{2}-\int_0^{2\pi} \frac{\cos(u)}{4}~du=\frac{\pi}{2}-\left.\frac{\sin(u)}{4}\right|_0^{2\pi}=\frac{\pi}{2}-\bigg(\frac{\sin(2\pi)}{4}-\frac{\sin(0)}{4}\bigg)=\frac{\pi}{2}.</math>
 
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Revision as of 15:44, 29 March 2016

Evaluate the indefinite and definite integrals.

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 \int \tan^3x ~dx}
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:  
Recall the trig identities:
1. 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. 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}}
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 \int \tan x~dx=\int \frac{-1}{u}~du=-\ln(u)+C=-\ln|\cos x|+C.}

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 \int \tan^3x~dx=\int (\sec^2x-1)\tan x ~dx=\int \sec^2x\tan x~dx-\int \tan x~dx.}
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
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 u~du-\int \tan x~dx=\frac{u^2}{2}-\int \tan x~dx=\frac{\tan^2x}{2}-\int \tan x~dx.}
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.} So, 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 \int \tan^3x~dx=\frac{\tan^2x}{2}+\int \frac{1}{u}~dx=\frac{\tan^2x}{2}+\ln |u|+C=\frac{\tan^2x}{2}+\ln |\cos x|+C.}

(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 \int_0^\pi \sin^2x~dx=\int_0^\pi \frac{1-\cos(2x)}{2}~dx=\int_0^\pi \frac{1}{2}~dx-\int_0^\pi \frac{\cos(2x)}{2}~dx.}
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 \int_0^\pi \sin^2x~dx=\left.\frac{x}{2}\right|_{0}^\pi-\int_0^\pi \frac{\cos(2x)}{2}~dx=\frac{\pi}{2}-\int_0^\pi \frac{\cos(2x)}{2}~dx.}
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 \int_0^\pi \sin^2x~dx=\frac{\pi}{2}-\int_0^{2\pi} \frac{\cos(u)}{4}~du=\frac{\pi}{2}-\left.\frac{\sin(u)}{4}\right|_0^{2\pi}=\frac{\pi}{2}-\bigg(\frac{\sin(2\pi)}{4}-\frac{\sin(0)}{4}\bigg)=\frac{\pi}{2}.}
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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