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jaersyn

  • 3 years ago

integrate (secx)^4 * sqrt (tanx) dx

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  1. Outkast3r09
    • 3 years ago
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    I think you have to use Trig sub

  2. Outkast3r09
    • 3 years ago
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    or you might be able to do this

  3. Outkast3r09
    • 3 years ago
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    this is just a test idk if it will work \[\int sec^2 x* sec^2 x*\sqrt{tanx} dx\]

  4. Outkast3r09
    • 3 years ago
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    \[cos^2 x + sin^2 x= 1\] \[1+tan^2 x=sec^2 x\]

  5. jaersyn
    • 3 years ago
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    you're correct, i'm close, but not exactly the ansewr

  6. Outkast3r09
    • 3 years ago
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    \[\int sec^2 x * (tan^2 x + 1) * \sqrt{tan x}dx\]

  7. jaersyn
    • 3 years ago
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    not trig sub, u-sub

  8. jaersyn
    • 3 years ago
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    yeah then let u = tanx; du = (secx)^2

  9. Outkast3r09
    • 3 years ago
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    you could use trig subs

  10. Outkast3r09
    • 3 years ago
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    trig sub will work with any trig integral... it's just a pain for something that can be done easier anyways \[\int sec^2 x *(tan^{3/2} x+ \sqrt{tan^{1/2}x})\]

  11. Outkast3r09
    • 3 years ago
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    \[\int sec^2 x*tan^{3/2}x dx + \int sec^2 x*tan^{1/2}xdx\]

  12. Outkast3r09
    • 3 years ago
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    \[\int sec^2 x* (tanx)^{3/2} dx + \int sec^2 x*(tanx)^{1/2}\]

  13. Outkast3r09
    • 3 years ago
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    let \[u=tanx\] \[du=sec^2 x\] \[\int u^{3/2}du+\int u^{1/2}du\]

  14. jaersyn
    • 3 years ago
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    i see where i went wrong

  15. Outkast3r09
    • 3 years ago
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    ok =]

  16. jaersyn
    • 3 years ago
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    ty

  17. jaersyn
    • 3 years ago
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    wait

  18. jaersyn
    • 3 years ago
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    ∫sec2x∗(tan3/2x+tan1/2x−−−−−−√)

  19. jaersyn
    • 3 years ago
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    how does tan^2 (x) * tan^(1/2) x = tan^(3/2) (x)?

  20. Outkast3r09
    • 3 years ago
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    ehh it's supposed to be a 5/2

  21. Outkast3r09
    • 3 years ago
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    so your answer wil lbe 7/2

  22. random231
    • 2 years ago
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    wow

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