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monroe17

  • 3 years ago

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

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  1. hartnn
    • 3 years ago
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    did you try tan 8x = u ?

  2. monroe17
    • 3 years ago
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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.

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

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

  5. anonymous
    • 3 years ago
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    turns in to \[\int\frac{u^2du}{\sqrt{16-u^2}}\]

  6. monroe17
    • 3 years ago
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    lol! okay I'll try to add brackets...

  7. anonymous
    • 3 years ago
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    then another sub \(z=\sin(4x)\) gives \[16\int \sin^2(z)dz\]

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

  9. hartnn
    • 3 years ago
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    did you mean z=4sin x ?

  10. monroe17
    • 3 years ago
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    added in brackets and i got it! thanks!

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

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