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

1. hartnn

did you try tan 8x = u ?

2. monroe17

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. satellite73

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

4. hartnn

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

5. satellite73

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

6. monroe17

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

7. satellite73

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

8. satellite73

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

9. hartnn

did you mean z=4sin x ?

10. monroe17

added in brackets and i got it! thanks!

11. hartnn

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