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

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!Step 1:    
 
!Step 1:    
 
|-
 
|-
−
|We start by writing <math>\int \tan^3xdx=\int \tan^2x\tan x dx</math>.  
+
|We start by writing <math>\int \tan^3x~dx=\int \tan^2x\tan x ~dx</math>.  
 
|-
 
|-
−
|Since <math>\tan^2x=\sec^2x-1</math>, we have <math>\int \tan^3xdx=\int (\sec^2x-1)\tan x dx=\int \sec^2\tan xdx-\int \tan xdx</math>.  
+
|Since <math>\tan^2x=\sec^2x-1</math>, we have <math>\int \tan^3x~dx=\int (\sec^2x-1)\tan x ~dx=\int \sec^2\tan x~dx-\int \tan x~dx</math>.  
 
|}
 
|}
  
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|Now, we need to use u substitution for the first integral. Let <math>u=\tan(x)</math>. Then, <math>du=\sec^2xdx</math>. So, we have
 
|Now, we need to use u substitution for the first integral. Let <math>u=\tan(x)</math>. Then, <math>du=\sec^2xdx</math>. So, we have
 
|-
 
|-
−
|<math>\int \tan^3xdx=\int udu-\int \tan xdx=\frac{u^2}{2}-\int \tan xdx=\frac{\tan^2x}{2}-\int \tan xdx</math>.
+
|<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;
 
|-
 
|-
−
|For the remaining integral, we need to use u substitution. First, we write <math>\int \tan^3xdx=\frac{\tan^2x}{2}-\int \frac{\sin x}{\cos x}dx</math>.
+
|For the remaining integral, we also need to use u substitution. First, we write <math>\int \tan^3x~dx=\frac{\tan^2x}{2}-\int \frac{\sin x}{\cos x}~dx</math>.
 
|-
 
|-
 
|Now, we let <math>u=\cos x</math>. Then, <math>du=-\sin xdx</math>. So, we get  
 
|Now, we let <math>u=\cos x</math>. Then, <math>du=-\sin xdx</math>. So, we get  
 
|-
 
|-
−
|<math>\int \tan^3xdx=\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>.
 
|}
 
|}
 
'''(b)'''
 
'''(b)'''
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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>.
 
|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>.
 
|-
 
|-
−
|Plugging this identity into our integral, we get <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>.  
+
|Plugging this identity into our integral, we get <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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|If we integrate the first integral, we get  
 
|If we integrate the first integral, we get  
 
|-
 
|-
−
|<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>.
+
|<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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|For the remaining integral, we need to use u substitution. Let <math>u=2x</math>. Then, <math>du=2dx</math>. Also, since this is a definite integral  
 
|For the remaining integral, we need to use u substitution. Let <math>u=2x</math>. Then, <math>du=2dx</math>. Also, since this is a definite integral  
 
|-
 
|-
−
|and we are using u substiution, 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>.
+
|and we are using u 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>.
 
|-
 
|-
 
|So, the integral becomes
 
|So, the integral becomes
 
|-
 
|-
−
|<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>
 
|}
 
|}
  

Revision as of 14:37, 31 January 2016

Evaluate the indefinite and definite integrals.


a)
b)


Foundations:  
Review u substitution
Trig identities

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^2\tan x~dx-\int \tan x~dx} .
Step 2:  
Now, we need to use 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^2xdx} . 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 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 xdx} . 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 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=2dx} . Also, since this is a definite integral
and we are using 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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