## inkyvoyd one year ago integrate sqrt(sqrt(tan x)) any way to do this?

1. anonymous

$\sqrt{\sqrt{tanx}} ???$

2. inkyvoyd

yup

3. anonymous

wow ok

4. inkyvoyd

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

5. inkyvoyd

so first, sqrt(sqrt(tan x))

6. anonymous

7. anonymous

this is messed up

8. inkyvoyd

but it's doable... how though

9. anonymous

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

10. inkyvoyd

it will only take 1 hour?

11. anonymous

r u asking me or telling me !

12. inkyvoyd

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

13. anonymous

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

14. inkyvoyd

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

15. anonymous

yes using substitution

16. anonymous

thats my approach

17. inkyvoyd

that's not specific enough...

18. Empty

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.

19. ganeshie8

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

20. ganeshie8

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

21. anonymous

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"...

22. anonymous

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

23. inkyvoyd

I finally see why the answer is so long.