Here's the question you clicked on:
monroe17
integral of (sec^2(8t)tan^2(8t))/(sqrt(16-tan^2(8t)))dt
did you try tan 8x = u ?
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.
i would guess putting \(u=\tan(x), du =\sec^2(x)dx\)
your this answer is also correct...i still think there is brackets mismatch somewhere, but couldn't find it :P
turns in to \[\int\frac{u^2du}{\sqrt{16-u^2}}\]
lol! okay I'll try to add brackets...
then another sub \(z=\sin(4x)\) gives \[16\int \sin^2(z)dz\]
which you can probably look up in a book or else rewrite in terms of \(\cos(2z)\)
added in brackets and i got it! thanks!
welcome ^_^ bdw, you did it on your own, so you deserve a medal, here it is :)