- #1

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anyone has an easy way to do the following problem..

[tex] \int \sqrt{\tan x} dx [/tex]

[tex] \int \sqrt{\tan x} dx [/tex]

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- Thread starter vincentchan
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- #1

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anyone has an easy way to do the following problem..

[tex] \int \sqrt{\tan x} dx [/tex]

[tex] \int \sqrt{\tan x} dx [/tex]

- #2

mathwonk

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= sec^2(x)dx. See if that makes it any better. you may to do some more substitutions later.

- #3

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http://integrals.wolfram.com/index.en.cgi

Try it out. Input Sqrt[Tan[x]]. Looks rather complicated :tongue2: .

- #4

Zurtex

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If [itex]du = \sec^2x dx[/itex] then:mathwonk said:

= sec^2(x)dx. See if that makes it any better. you may to do some more substitutions later.

[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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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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then dx/dt=sec(square)t

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

and use partial fractions ...thats it..

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