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monroe17

integral of (sec^2(8t)tan^2(8t))/(sqrt(16-tan^2(8t)))dt

  • one year ago
  • one year ago

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  1. hartnn
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    did you try tan 8x = u ?

    • one year ago
  2. monroe17
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    I got.. (sec(8t)(16sqrt(34cos(16t)+30))arctan((sqrt(2)sin(8t))/(sqrt(17cos(16t)+15))-(17cos(16t)+15)tan(8t)sec(8t)))/(32sqrt(16-tan^2(8t))) but it says its wrong.. once again.

    • one year ago
  3. satellite73
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    i would guess putting \(u=\tan(x), du =\sec^2(x)dx\)

    • one year ago
  4. hartnn
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    your this answer is also correct...i still think there is brackets mismatch somewhere, but couldn't find it :P

    • one year ago
  5. satellite73
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    turns in to \[\int\frac{u^2du}{\sqrt{16-u^2}}\]

    • one year ago
  6. monroe17
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    lol! okay I'll try to add brackets...

    • one year ago
  7. satellite73
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    then another sub \(z=\sin(4x)\) gives \[16\int \sin^2(z)dz\]

    • one year ago
  8. satellite73
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    which you can probably look up in a book or else rewrite in terms of \(\cos(2z)\)

    • one year ago
  9. hartnn
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    did you mean z=4sin x ?

    • one year ago
  10. monroe17
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    added in brackets and i got it! thanks!

    • one year ago
  11. hartnn
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    welcome ^_^ bdw, you did it on your own, so you deserve a medal, here it is :)

    • one year ago
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