Difference between revisions of "009B Sample Final 2, Problem 6"
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|Let <math>x=4\sec \theta.</math> | |Let <math>x=4\sec \theta.</math> | ||
|- | |- | ||
| − | |Then, <math>dx=4\sec \theta \tan \theta d\theta.</math> | + | |Then, <math>dx=4\sec \theta \tan \theta ~d\theta.</math> |
|- | |- | ||
|So, the integral becomes | |So, the integral becomes | ||
| Line 83: | Line 83: | ||
|Now, we use <math>u</math>-substitution. | |Now, we use <math>u</math>-substitution. | ||
|- | |- | ||
| − | |Let <math>u=\sin x.</math> Then, <math>du=\cos x dx.</math> | + | |Let <math>u=\sin x.</math> Then, <math>du=\cos x ~dx.</math> |
|- | |- | ||
|Since this is a definite integral, we need to change the bounds of integration. | |Since this is a definite integral, we need to change the bounds of integration. | ||
| Line 113: | Line 113: | ||
| <math>\frac{x-3}{(x+1)(x+5)}=\frac{A}{x+1}+\frac{B}{x+5}.</math> | | <math>\frac{x-3}{(x+1)(x+5)}=\frac{A}{x+1}+\frac{B}{x+5}.</math> | ||
|- | |- | ||
| − | |If we multiply both sides of this equation by <math>(x+1)(x+5),</math> we get | + | |If we multiply both sides of this equation by <math>(x+1)(x+5),</math> we get |
|- | |- | ||
| <math>x-3=A(x+5)+B(x+1).</math> | | <math>x-3=A(x+5)+B(x+1).</math> | ||
|- | |- | ||
| − | |If we let <math>x=-1,</math> we get <math>A=-1.</math> | + | |If we let <math>x=-1,</math> we get <math>A=-1.</math> |
|- | |- | ||
| − | |If we let <math>x=-5,</math> we get <math>B=2.</math> | + | |If we let <math>x=-5,</math> we get <math>B=2.</math> |
|- | |- | ||
|So, we have | |So, we have | ||
Revision as of 13:16, 4 March 2017
Evaluate the following 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 \frac{dx}{x^2\sqrt{x^2-16}}}
(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_{-\pi}^\pi \sin^3x\cos^3x~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_0^1 \frac{x-3}{x^2+6x+5}~dx}
| Foundations: |
|---|
| 1. 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 \int \frac{dx}{x^2\sqrt{x^2-16}},} what would be the correct trig substitution? |
| The correct substitution 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 x=4\sec^2\theta.} |
| 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 \cos^2(x)=1-\sin^2(x).} |
| 3. 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.} |
Solution:
(a)
| Step 1: |
|---|
| We start by using trig 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 x=4\sec \theta.} |
| 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 dx=4\sec \theta \tan \theta ~d\theta.} |
| 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 \frac{1}{x^2\sqrt{x^2-16}}~dx} & = & \displaystyle{\int \frac{4\sec \theta \tan \theta}{16\sec^2\theta \sqrt{16\sec^2 \theta -16}}~d\theta}\\ &&\\ & = & \displaystyle{\int \frac{4\sec \theta \tan \theta}{16\sec^2\theta (4\tan \theta)} ~d\theta}\\ &&\\ & = & \displaystyle{\int \frac{1}{16\sec \theta} ~d\theta.} \end{array}} |
| Step 2: |
|---|
| Now, we integrate 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 \begin{array}{rcl} \displaystyle{\int \frac{1}{x^2\sqrt{x^2-16}}~dx} & = & \displaystyle{\frac{1}{16}\cos\theta~d\theta}\\ &&\\ & = & \displaystyle{\frac{1}{16}\sin \theta +C}\\ &&\\ & = & \displaystyle{\frac{1}{16}\bigg(\frac{\sqrt{x^2-16}}{x}\bigg)+C.} \end{array}} |
(b)
| 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 \begin{array}{rcl} \displaystyle{\int_{-\pi}^\pi \sin^3x\cos^3x~dx} & = & \displaystyle{\int_{-\pi}^{\pi} \sin^3x \cos^2x \cos x~dx}\\ &&\\ & = & \displaystyle{\int_{-\pi}^{\pi} \sin^3x (1-\sin^2x)\cos x~dx.} \end{array}} |
| Step 2: |
|---|
| Now, 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=\sin 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=\cos x ~dx.} |
| Since this is a definite integral, we need to change the bounds of integration. |
| Then, 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=\sin(-\pi)=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=\sin(\pi)=0.} |
| 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_{-\pi}^\pi \sin^3x\cos^3x~dx} & = & \displaystyle{\int_0^0 u^3(1-u^2)~du}\\ &&\\ & = & \displaystyle{0.} \end{array}} |
(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_0^1 \frac{x-3}{x^2+6x+5}~dx=\int_0^1 \frac{x-3}{(x+1)(x+5)}~dx.} |
| Now, we use partial fraction decomposition. Wet set |
| 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{x-3}{(x+1)(x+5)}=\frac{A}{x+1}+\frac{B}{x+5}.} |
| If we multiply both sides of this 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+1)(x+5),} 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 x-3=A(x+5)+B(x+1).} |
| 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=-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 A=-1.} |
| 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=-5,} 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 B=2.} |
| 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 \frac{x-3}{(x+1)(x+5)}=\frac{-1}{x+1}+\frac{2}{x+5}.} |
| Step 2: |
|---|
| Now, 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_0^1 \frac{x-3}{(x+1)(x+5)}~dx} & = & \displaystyle{\int_0^1 \frac{-1}{x+1}+\frac{2}{x+5}~dx}\\ &&\\ & = & \displaystyle{\int_0^1 \frac{-1}{x+1}~dx+\int_0^1 \frac{2}{x+5}~dx.} \end{array}} |
| Now, 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 for both of these integrals. |
| 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+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=dx.} |
| 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 t=x+5.} 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 dt=dx.} |
| Since these are definite integrals, 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=0+1=1} 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=1+1=2.} |
| Also, 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 t_1=0+5=5} 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=1+5=6.} |
| 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_0^1 \frac{x-3}{(x+1)(x+5)}~dx} & = & \displaystyle{\int_1^2 \frac{-1}{u}~du+\int_5^6 \frac{2}{t}~dt}\\ &&\\ & = & \displaystyle{-\ln|u|\bigg|_1^2+2\ln|t|\bigg|_5^6}\\ &&\\ & = & \displaystyle{-\ln(2)+2\ln(6)-2\ln(5).} \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{1}{16}\bigg(\frac{\sqrt{x^2-16}}{x}\bigg)+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 0} |
| (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 -\ln(2)+2\ln(6)-2\ln(5)} |