Question

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

Answer 1

did you try tan 8x = u ?

Answer 2

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.

Answer 3

i would guess putting
\(u=\tan(x), du =\sec^2(x)dx\)

Answer 4

your this answer is also correct...i still think there is brackets mismatch somewhere, but couldn't find it :P

Answer 5

turns in to
\[\int\frac{u^2du}{\sqrt{16-u^2}}\]

Answer 6

lol! okay I'll try to add brackets...

Answer 7

then another sub \(z=\sin(4x)\) gives
\[16\int \sin^2(z)dz\]

Answer 8

which you can probably look up in a book or else rewrite in terms of \(\cos(2z)\)

Answer 9

did you mean z=4sin x ?

Answer 10

added in brackets and i got it! thanks!

Answer 11

welcome ^_^
bdw, you did it on your own, so you deserve a medal, here it is :)