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monroe17
 3 years ago
integral of (sec^2(8t)tan^2(8t))/(sqrt(16tan^2(8t)))dt
monroe17
 3 years ago
integral of (sec^2(8t)tan^2(8t))/(sqrt(16tan^2(8t)))dt

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hartnn
 3 years ago
Best ResponseYou've already chosen the best response.1did you try tan 8x = u ?

monroe17
 3 years ago
Best ResponseYou've already chosen the best response.1I 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.

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.0i would guess putting \(u=\tan(x), du =\sec^2(x)dx\)

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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.0turns in to \[\int\frac{u^2du}{\sqrt{16u^2}}\]

monroe17
 3 years ago
Best ResponseYou've already chosen the best response.1lol! okay I'll try to add brackets...

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.0then another sub \(z=\sin(4x)\) gives \[16\int \sin^2(z)dz\]

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.0which you can probably look up in a book or else rewrite in terms of \(\cos(2z)\)

monroe17
 3 years ago
Best ResponseYou've already chosen the best response.1added in brackets and i got it! thanks!

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