inkyvoyd
  • inkyvoyd
integrate sqrt(sqrt(tan x)) any way to do this?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\sqrt{\sqrt{tanx}} ???\]
inkyvoyd
  • inkyvoyd
yup
anonymous
  • anonymous
wow ok

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inkyvoyd
  • 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
inkyvoyd
  • inkyvoyd
so first, sqrt(sqrt(tan x))
anonymous
  • anonymous
anonymous
  • anonymous
this is messed up
inkyvoyd
  • inkyvoyd
but it's doable... how though
anonymous
  • anonymous
did u see the attachment its doable if u have an free hour in ur hand
inkyvoyd
  • inkyvoyd
it will only take 1 hour?
anonymous
  • anonymous
r u asking me or telling me !
inkyvoyd
  • inkyvoyd
I dont' know... just wonder if it's doable
anonymous
  • anonymous
it is doable yea , as u can see wolfram puked the answer out however it ll take alot of time
inkyvoyd
  • inkyvoyd
I don't mind... but do you know how to start it?
anonymous
  • anonymous
yes using substitution
anonymous
  • anonymous
thats my approach
inkyvoyd
  • inkyvoyd
that's not specific enough...
Empty
  • 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.
ganeshie8
  • 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
ganeshie8
  • ganeshie8
you could turn it into a beta integral if it is a definite integral between 0->1
anonymous
  • 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"...
anonymous
  • anonymous
Meh, I wouldn't count on IBP here...
inkyvoyd
  • inkyvoyd
I finally see why the answer is so long.

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