Calculus problem

  • #1
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anyone has an easy way to do the following problem..
[tex] \int \sqrt{\tan x} dx [/tex]
 

Answers and Replies

  • #2
mathwonk
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the basic idea of substitution is to try to put a new symbol in for anything you find confusing. Like here you might try u = tan(x), and then of course you have to put du
= sec^2(x)dx. See if that makes it any better. you may to do some more substitutions later.
 
  • #4
Zurtex
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mathwonk said:
the basic idea of substitution is to try to put a new symbol in for anything you find confusing. Like here you might try u = tan(x), and then of course you have to put du
= sec^2(x)dx. See if that makes it any better. you may to do some more substitutions later.
If [itex]du = \sec^2x dx[/itex] then:

[tex]\frac{du}{1 + u^2} = dx[/tex]

Hope that helps. Your integral is now:

[tex]\int \frac{\sqrt{u}}{1 + u^2}du[/tex]

I'd look at using byparts from there on, but that's just a guess.
 
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  • #5
arildno
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parth dave's substitution seems best to me.
We have:
[tex]u=\sqrt{tan{x}}\to{du}=\frac{dx}{2u}\frac{1}{\cos^{2}x}=\frac{dx}{2u}(u^{4}+1)[/tex]
Or:
[tex]\int\sqrt{tan{x}}dx=\int\frac{2u^{2}}{u^{4}+1}du[/tex]

We note the identity:
[tex]u^{4}+1=(u^{2}-\sqrt{2}u+1)(u^{2}+\sqrt{2}u+1)[/tex]

We may now use partial fractions techniques to derive the answer.
 
  • #6
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Put tanx=t square

then dx/dt=sec(square)t

put sec(sqr)t= 1+t(sqr)

and use partial fractions ...thats it..
 

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