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

Revision as of 09:23, 7 February 2017

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:
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
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 u~du-\int \tan x~dx}\\ &&\\ & = & \displaystyle{\frac{u^2}{2}-\int \tan x~dx}\\ &&\\ & = & \displaystyle{\frac{\tan^2x}{2}-\int \tan x~dx.}\\ \end{array}}

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}}

Return to Sample Exam

@@ Line 8: / Line 8: @@
 !Foundations: &nbsp;
 |-
-|Recall the trig identities:
+|'''1.''' Recall the trig identity
 |-
-|'''1.''' <math style="vertical-align: -3px">\tan^2x+1=\sec^2x</math>
+|&nbsp; &nbsp; <math style="vertical-align: -3px">\tan^2x+1=\sec^2x</math>
 |-
-|'''2.''' <math style="vertical-align: -13px">\sin^2(x)=\frac{1-\cos(2x)}{2}</math>
+|'''2.''' Recall the trig identity
+|-
+|&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>
 |-
 |
-::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; 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>
 |-
 |
-::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; Now, let <math style="vertical-align: -5px">u=\cos(x).</math> Then, <math style="vertical-align: -5px">du=-\sin(x)dx.</math> Thus,
 |-
 |
-::<math>\begin{array}{rcl}
+&nbsp; &nbsp; <math>\begin{array}{rcl}
 \displaystyle{\int \tan x~dx} & = & \displaystyle{\int \frac{-1}{u}~du}\\
 &&\\
@@ Line 82: / Line 84: @@
 &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> So, we get
+|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
 |-
 |
@@ Line 100: / Line 104: @@
 |One of the double angle formulas is <math style="vertical-align: -5px">\cos(2x)=1-2\sin^2(x).</math>
 |-
-|Solving for <math style="vertical-align: -5px">\sin^2(x),</math> we get <math style="vertical-align: -13px">\sin^2(x)=\frac{1-\cos(2x)}{2}.</math>
+|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>
 |-
 |Plugging this identity into our integral, we get
@@ Line 132: / Line 138: @@
 |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, we need to change the bounds of integration.
+|Also, since this is a definite integral and we are using <math style="vertical-align: 0px">u</math>-substitution,
+|-
+|we need to change the bounds of integration.
 |-
 |We have <math style="vertical-align: -5px">u_1=2(0)=0</math> and <math style="vertical-align: -5px">u_2=2(\pi)=2\pi.</math>

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

Revision as of 09:23, 7 February 2017

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