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hartnnBest ResponseYou've already chosen the best response.1
did you try tan 8x = u ?
 one year ago

monroe17Best ResponseYou've already chosen the best response.1
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(16tan^2(8t))) but it says its wrong.. once again.
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
i would guess putting \(u=\tan(x), du =\sec^2(x)dx\)
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
your this answer is also correct...i still think there is brackets mismatch somewhere, but couldn't find it :P
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
turns in to \[\int\frac{u^2du}{\sqrt{16u^2}}\]
 one year ago

monroe17Best ResponseYou've already chosen the best response.1
lol! okay I'll try to add brackets...
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
then another sub \(z=\sin(4x)\) gives \[16\int \sin^2(z)dz\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
which you can probably look up in a book or else rewrite in terms of \(\cos(2z)\)
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
did you mean z=4sin x ?
 one year ago

monroe17Best ResponseYou've already chosen the best response.1
added in brackets and i got it! thanks!
 one year ago

hartnnBest ResponseYou've already chosen the best response.1
welcome ^_^ bdw, you did it on your own, so you deserve a medal, here it is :)
 one year ago
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