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Integration With Trig Substitution Calc II

  • Thread starter Wm_Davies
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  • #1
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Homework Statement




[tex]\int \frac{\sqrt{196 x^2-144}}{x} dx[/tex]


Homework Equations





The Attempt at a Solution




I first rewrote the integral.....
[tex]\int \frac{\sqrt{(14x)^2-12^2}}{x} dx[/tex]
Then I let.......
14x=12sec[tex]\theta[/tex]
thus....
x=6/7sec[tex]\theta[/tex]
dx=6/7sec[tex] \theta tan \theta d \theta[/tex]

My progression of this problem is as follows:

[tex] \int \frac{\sqrt{(144(sec^2 \theta -1)}}{6/7sec \theta} d \theta[/tex]
[tex]12 \int \frac{tan \theta}{6/7sec \theta}(6/7sec \theta tan \theta) d \theta[/tex]
[tex]12 \int tan^2 \theta d \theta[/tex]
[tex]12 \int sec^2 - 1 \theta d \theta[/tex]
[tex]12 tan \theta [/tex]

I am not sure how to finish the problem after that. I am also not sure I did that right since I am very new at this, but I have a feeling everything I did was right.
 

Answers and Replies

  • #2
LCKurtz
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I didn't check all your steps, but didn't you forget to integrate the -θ in the last integral?

Now draw a little right triangle in the first quadrant with hypotenuse 7x, x-side of 6 and y-side of [itex]\sqrt{(49x^2-36)}[/itex] and polar angle θ. You should be able to read everything off this triangle to get your answer back in terms of x.
 
  • #3
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I didn't check all your steps, but didn't you forget to integrate the -θ in the last integral?
Sorry the last integral was a typo it should be....

[tex]12 \int sec^2 \theta - 1 d \theta[/tex]


Okay so after I fill this in the answer should be....

[tex]\frac{\sqrt{49x^2-36}}{6}+c[/tex]
 
  • #4
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That answer is wrong. I am not sure what mistake I made.
 
  • #5
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That answer is wrong. I am not sure what mistake I made.
I forgot to multiply it by 12 so I would have had...

[tex]2 \sqrt{49x^2 - 36} + c [/tex]

which is still wrong

I checked Wolfram Alpha and they gave me the answer as.....

[tex]2 \sqrt{49 x^2 - 36} + 12 tan^-1\left[\frac {6}{\sqrt{49 x^2 -36}}\right] + c[/tex]

So I now know that I am close, but after checking it with the steps in Wolfram Alpha I am still not sure where I went wrong.
 
Last edited:
  • #6
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What is ∫tan2θ dθ? There's really no need to change tan2θ to sec2θ - 1 that I can see.
Once you find that integral, make sure you substitute for x correctly.
 
  • #7
vela
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So I now know that I am close, but after checking it with the steps in Wolfram Alpha I am still not sure where I went wrong.
You forgot to integrate the 1 in the integrand.
 
  • #8
51
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You forgot to integrate the 1 in the integrand.
Wow, I did forget to do that! Thanks for noticing that. I got it now.
 

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