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inkyvoyd

  • one year ago

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

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  1. anonymous
    • one year ago
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    \[\sqrt{\sqrt{tanx}} ???\]

  2. inkyvoyd
    • one year ago
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    yup

  3. anonymous
    • one year ago
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    wow ok

  4. inkyvoyd
    • one year ago
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    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
    • one year ago
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    so first, sqrt(sqrt(tan x))

  6. anonymous
    • one year ago
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  7. anonymous
    • one year ago
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    this is messed up

  8. inkyvoyd
    • one year ago
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    but it's doable... how though

  9. anonymous
    • one year ago
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    did u see the attachment its doable if u have an free hour in ur hand

  10. inkyvoyd
    • one year ago
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    it will only take 1 hour?

  11. anonymous
    • one year ago
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    r u asking me or telling me !

  12. inkyvoyd
    • one year ago
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    I dont' know... just wonder if it's doable

  13. anonymous
    • one year ago
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    it is doable yea , as u can see wolfram puked the answer out however it ll take alot of time

  14. inkyvoyd
    • one year ago
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    I don't mind... but do you know how to start it?

  15. anonymous
    • one year ago
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    yes using substitution

  16. anonymous
    • one year ago
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    thats my approach

  17. inkyvoyd
    • one year ago
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    that's not specific enough...

  18. Empty
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    you could turn it into a beta integral if it is a definite integral between 0->1

  21. anonymous
    • one year ago
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    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
    • one year ago
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    Meh, I wouldn't count on IBP here...

  23. inkyvoyd
    • one year ago
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    I finally see why the answer is so long.

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