Question

@bmp ,
how'd you integrate? sqrt tan x? Small hint, smaller than telling me what u is, please.

Answer 1

Haha, trying that, mate? What did you get after making u = tan(x) ?

Answer 2

eh, and multiplying by (sec^2 x-tan^2 x)?

Answer 3

Got a never ending chain, thanks a lot :S

Answer 4

Actually, you should end up with something like \[ \int \sqrt{\tan{x}}dx = \int \frac{\sqrt{u}}{u^2 + 1}du \]And then it's a never ending chain of partial fractions and u-subs.

Answer 5

What?

Answer 6

:S

Answer 7

If u = tan(x), du = sec^2(x)dx, right? Then: \[dx = \frac{du}{sec^2{x}} \]Rewrite the sec^2 as tan and substitute it :-)

Answer 8

I'm going to cheat now. *wolfram*

Answer 9

Go ahead. :-) I don't think a sane person should work this integral out; or, do it only once in a life time :-).

Answer 10

That is one really complicated expression,

Answer 11

That is one really complicated show steps.

Answer 12

Haha. Yeah, it's. It's very messy. Not that complicated, tho. Just a bunch of partial fractions and algebra.

Answer 13

Did you do ever do that?
and I know why I couldn't do it. I haven't learned rational integration (I do remember skimming it and seeing the substition to prevent infinite chain)

Answer 14

I did, but like I said, I will never do it again. It was towards the end of my Calc I class. I almost failed my integral exam because I spend two nights on this. But, well, it was more or less worth it.

Answer 15

I should've hinted that also for you :-) My bad.

Answer 16

After I learn partial fractions, I'm going to do this >:)

Answer 17

:-D. It's interesting. Also, a correction from yesterday: \[ \int_{0}^{\infty} sinx^2dx = \frac{1}{2} \sqrt{\frac{\pi}{2}} \]Not true in general.

Answer 18

Anyway, gotta sleep now, mate. Cya.

Answer 19

eh?
alright, bb