Question

integrate sqrt(sqrt(tan x))
any way to do this?

Answer 1

\[\sqrt{\sqrt{tanx}} ???\]

Answer 2

yup

Answer 3

wow ok

Answer 4

I know how to do sqrt(tan x). it's very messy. but I wonder how to do (tan x)^(1/(2n)) for all natural numbers n

Answer 5

so first, sqrt(sqrt(tan x))

Answer 6

this is messed up

Answer 7

but it's doable... how though

Answer 8

did u see the attachment its doable if u have an free hour in ur hand

Answer 9

it will only take 1 hour?

Answer 10

r u asking me or telling me !

Answer 11

I dont' know... just wonder if it's doable

Answer 12

it is doable yea , as u can see wolfram puked the answer out however it ll take alot of time

Answer 13

I don't mind... but do you know how to start it?

Answer 14

yes using substitution

Answer 15

thats my approach

Answer 16

that's not specific enough...

Answer 17

I think you can probably rewrite it as \[\int \frac{\sin^{1/4}x}{\cos^{1/4}x} x dx\]
And then from there I would try to screw around with some substitutions maybe, I am only kinda guessing this because there ended up being some logarithms and garbage.

Answer 18

subing \(u^4 = \tan x\) gives
\[\int \dfrac{4u^4}{1+u^8}\,du\]
which ofcourse is a pain even for those who like partial fractions so much

Answer 19

you could turn it into a beta integral if it is a definite integral between 0->1

Answer 20

Hmm...
\[\begin{align*}\frac{u^4}{u^8+1}&=\frac{u^4}{2}\left(\frac{u^4+1}{u^8+1}-\frac{u^4-1}{u^8+1}\right)\\[2ex]
&=\frac{u^4}{2}\left(\frac{1+\dfrac{1}{u^4}}{u^4+\dfrac{1}{u^4}}-\frac{1-\dfrac{1}{u^4}}{u^4+\dfrac{1}{u^4}}\right)\\[2ex]
&=\frac{u^4}{2}\left(\frac{1+\dfrac{1}{u^4}}{\left(u^2-\dfrac{1}{u^2}\right)^2+2}-\frac{1-\dfrac{1}{u^4}}{\left(u^2+\dfrac{1}{u^2}\right)^2-2}\right)\\[2ex]
&=\frac{u^3}{4}\left(\frac{2u+\dfrac{2}{u^3}}{\left(u^2-\dfrac{1}{u^2}\right)^2+2}-\frac{2u-\dfrac{2}{u^3}}{\left(u^2+\dfrac{1}{u^2}\right)^2-2}\right)\end{align*}\]
Maybe we can integrate by parts? The rational terms to the right can be integrated "nicely"...

Answer 21

Meh, I wouldn't count on IBP here...

Answer 22

I finally see why the answer is so long.