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Inverse Trig Substitution Integral

  • Thread starter Bazzinga
  • Start date
  • #1
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[tex]\int x^{2} \sqrt{4+x^{2}} dx[/tex]

I've already subbed in:

[tex] x = 2tan\theta [/tex]
[tex] dx = 2sec^{2}\theta d\theta [/tex]

and I've gotten down to:

[tex] 16 \int tan^{3}\theta sec^{3}\theta d\theta [/tex]

But now I have noo idea what to do! Can someone give me a hint?
 

Answers and Replies

  • #2
tiny-tim
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Hi Bazzinga! :smile:

(where did your tan3 come from? :confused:)

sorry, but you just have to slog it out, using repeated integration by parts :redface:

substituting x = sinh(u) instead would have been a lot easier :wink:
 
  • #3
rock.freak667
Homework Helper
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I think your integral should reduce to
[tex]
16 \int tan^{2}\theta sec^{3}\theta d\theta
[/tex]

as the x2 term outside the square root will give tan2.

Replace tan2θ with sec2θ -1 and try to simplify it. I think the integrals may get a bit complicated.
 
  • #4
SammyS
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...
and I've gotten down to:

[tex] 16 \int \tan^{3}\theta \sec^{3}\theta d\theta [/tex]
I think your integral should reduce to
[tex]
16 \int \tan^{2}\theta \sec^{3}\theta d\theta
[/tex]

as the x2 term outside the square root will give tan2.

Replace tan2θ with sec2θ -1 and try to simplify it. I think the integrals may get a bit complicated.
Hi Bazzinga.

The integral:
[tex] 16 \int \tan^{3}\theta \sec^{3}\theta \,d\theta [/tex]
works out very easily with rock.freak667's suggestion for a replacement.

However, if rock.freak667 is right (probably the case) then your integral reduces to
[tex]
16 \int \sec^{5}\theta -\sec^{3}\theta d\theta
[/tex]
 
  • #5
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Oh sorry, the term is [tex]x^{3}[/tex] not [tex]x^{2}[/tex]
 
  • #6
SammyS
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Oh sorry, the term is [tex]x^{3}[/tex] not [tex]x^{2}[/tex]
In that case,
[tex]
\int \tan^{3}\theta \sec^{3}\theta \,d\theta = \int \tan\theta\,\left(\sec^{5}\theta-\sec^{3}\theta \right)\, d\theta
[/tex]
[tex]=\int \left(\sec^{4}\theta-\sec^{2}\theta \right)\tan\theta\,\sec\theta\ d\theta[/tex]
This is quite straight forward.
 

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