calculus problem

[vincentchan]

anyone has an easy way to do the following problem..
\int \sqrt{\tan x} dx

[mathwonk]

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.

[Parth Dave]

I don't think a substitution is going to do it.

http://integrals.wolfram.com/index.en.cgi
Try it out. Input Sqrt[Tan[x]]. Looks rather complicated :tongue2: .

[Zurtex]

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 du = \sec^2x dx then:

\frac{du}{1 + u^2} = dx

Hope that helps. Your integral is now:

\int \frac{\sqrt{u}}{1 + u^2}du

I'd look at using byparts from there on, but that's just a guess.

[arildno]

parth dave's substitution seems best to me.
We have:
u=\sqrt{tan{x}}\to{du}=\frac{dx}{2u}\frac{1}{\cos^ {2}x}=\frac{dx}{2u}(u^{4}+1)
Or:
\int\sqrt{tan{x}}dx=\int\frac{2u^{2}}{u^{4}+1}du

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

We may now use partial fractions techniques to derive the answer.

[Dr.Brain]

Put tanx=t square

then dx/dt=sec(square)t

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

and use partial fractions ...thats it..