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Mentor
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posted by  Vinisgood on 10/31/2009 1:57:18 AM  |  status: Closed  |  Earned Karma: 561

HELP DUE IN 1 HOUR!!!

Course Textbook Chapter Problem Needs by
Calculus N/A N/A N/A 10/31/2009 at 2:00:00 PM
Question Details:
evaluate:
int (dt)/sqrt(t^2-8 t + 25)


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Oracle
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posted by hado on 10/31/2009 2:12:44 AM  |  status: Live
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Response Details:
First, complete the square




Let

==>


 
Second method,
First, complete the square


Let

Apply the formula

==>

==>

==>


 
Both answers are the same
Sage
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posted by Edward E. on 10/31/2009 2:13:07 AM  |  status: Live
Asker's Rating: Helpful   
Response Details:
Using the integral table we have:



We can rewrite the integral in this format:
Oracle
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posted by anonymousxyz on 10/31/2009 2:29:57 AM  |  status: Live
Asker's Rating: Helpful   
Response Details:


\begin{align*} & \int \frac{\text{d}t}{\sqrt{t^2 - 8t + 25}} &=\\\\& \int \frac{\text{d}t}{\sqrt{t^2 - 8t + 16 - 16 + 25}} &=\\\\& \int \frac{\text{d}t}{\sqrt{(t - 4)^2 + 9}} \end{align*}


t = 3 \tan \theta + 4, \qquad \frac{\text{d}t}{\text{d}\theta} = 3 \sec^2 \theta, \qquad \text{d}t = 3 \sec^2 \theta \; \text{d}\theta


\begin{align*} & \int \frac{\text{d}t}{\sqrt{(t - 4)^2 + 9}} &=\\\\& \int \frac{3\sec^2 \theta}{\sqrt{9 \tan^2 \theta + 9}}\;\text{d}\theta &=\\\\& \int \frac{3\sec^2 \theta}{\sqrt{9 (\tan^2 \theta + 1 )}}\;\text{d}\theta &=\\\\& \int \frac{3\sec^2 \theta}{\sqrt{9\sec^2 \theta}}\;\text{d}\theta &=\\\\& \int \sec \theta\;\text{d}\theta &=\\\\& \ln | \sec \theta + \tan \theta | + C \end{align*}


\begin{align*} & \text{Think of a right triangle:} \\\\& t = 3 \tan \theta + 4 \qquad \Rightarrow \qquad \tan \theta = \frac{t - 4}{3} = \frac{\text{opp}}{\text{adj}} \\\\& \text{hyp} = \sqrt{\text{opp}^2 + \text{adj}^2} = \sqrt{ (t - 4)^2 + 9 } \\\\& \cos \theta = \frac{\text{adj}}{\text{hyp}} = \frac{3}{\sqrt{(t - 4)^2 + 9}} \qquad \Rightarrow \qquad \sec\theta = \frac{1}{\cos \theta} = \frac{\sqrt{(t - 4)^2 + 9}}{3} \end{align*}


\begin{align*} & \ln | \sec \theta + \tan \theta | + C &=\\\\& \ln \left| \frac{\sqrt{(t - 4)^2 + 9}}{3} + \frac{t - 4}{3} \right| + C &=\\\\& \sinh^{-1}\left( \frac{t-4}{3}\right) + C \end{align*}
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