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Applying Integration by Parts & Trig Substitution

  • Thread starter wvcaudill2
  • Start date
  • #1
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Homework Statement



[tex]\int\sqrt{4+9x^{2}}dx[/tex]


Homework Equations


Pythagorean Identities?



The Attempt at a Solution


I find it sort of cumbersome to use the special formatting here, so I hope it is okay that I just photocopied my work on paper.

You can see how far I made it, but now I am stuck. I am pretty sure I need to use integration by parts to finish the problem, but I am not sure how.
scan0001.jpg

Sorry about all the erasure marks, they showed up kind of dark.
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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So all you need to do is have

u=secθ and dv=sec2θ dθ

and you know that ∫ u dv = uv - ∫ v du.
 
  • #3
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So all you need to do is have

u=secθ and dv=sec2θ dθ

and you know that ∫ u dv = uv - ∫ v du.

How do I integrate dv? I know the integral of sec θ = ln ( sec θ + tan θ) + c
But what about sec2?
 
  • #4
Matterwave
Science Advisor
Gold Member
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The integral of sec^2(x) is tan(x)+C. You can figure this out by simply noting that the derivative of tan=sin/cos is sec^2.
 
  • #5
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The integral of sec^2(x) is tan(x)+C. You can figure this out by simply noting that the derivative of tan=sin/cos is sec^2.
Wow, I can't believe I missed this! Thank you!
 
  • #6
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Alright, I still seem to have a problem. I now have
[tex](3/4)(secθtanθ)-\int tan^{2}θsecθ[/tex]

Where do I go from here?
 
Last edited:
  • #7
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Is there anyone that can help me finish this problem?
 
  • #8
rock.freak667
Homework Helper
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Is there anyone that can help me finish this problem?
you'll need to put back in 1+tan2θ = sec2θ (yes this will give you back an integral of sec3θ)

Now remember you got something like

∫sec3θ dθ = secθ tanθ - ∫tan2θsecθ dθ

you'll see that you can put the ∫sec3θ dθ (that you get from the substitution) back on the left side. If you don't get what I mean, write it all out and you should hopefully see it.
 
  • #9
54
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I am still not seeing it.

Currently, I have
[tex]\int sec^{3}Θ = (3/4)(secΘtanΘ - \int tan^{2}ΘsecΘ) [/tex]
 
  • #10
rock.freak667
Homework Helper
6,230
31
I am still not seeing it.

Currently, I have
[tex]\int sec^{3}Θ = (3/4)(secΘtanΘ - \int tan^{2}ΘsecΘ) [/tex]

So when you put in [itex]tan^2 \theta = sec^2\theta -1[/itex]

[tex]\int sec^3 \theta d\theta = \frac{3}{4}sec\theta tan\theta - \int sec\theta (sec^2\theta-1) d\theta[/tex]

[tex] \int sec^3 \theta d\theta = \frac{3}{4}sec\theta tan\theta - \int sec^3 \theta d\theta -\int sec \theta d\theta[/tex]

So you can move that ∫ sec3θ dθ to the left side to get 2∫ sec3θ dθ.
 
  • #11
54
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Okay, thats what I thought you wanted me to do. I am now left with this:

[tex](7/4)\int sec^{3}Θ = (3/4)secΘtanΘ - (3/4)ln\left|secΘ + tanΘ\right| + C[/tex]

Now I need to manipulate the fraction so as to get
[tex](4/3)\int sec^{3}Θ[/tex]

How do I do this?
 
  • #12
rock.freak667
Homework Helper
6,230
31
Okay, thats what I thought you wanted me to do. I am now left with this:

[tex](7/4)\int sec^{3}Θ = (3/4)secΘtanΘ - (3/4)ln\left|secΘ + tanΘ\right| + C[/tex]

Now I need to manipulate the fraction so as to get
[tex](4/3)\int sec^{3}Θ[/tex]

How do I do this?
multiply both sides by 4/7 and then multiply by 4/3. The left side would give ∫ (4/3)sec3θ dθ.
 
  • #13
54
0
Okay, I got it now. Thanks for all your help!
 

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