Question

Trig integration: errrg

Answer 1

oo the fun stuff

Answer 2

\[\int\limits_{}^{}8tanxsec^3xdx\]

Answer 3

I tried factoring out one sec to get sec^2x but

Answer 4

and use the pythagorean identity but it didnt work

Answer 5

Maybe there is a simple u-sub somewhere,
(tan x)' =
(sec x)' =

remember these?

Answer 6

\[ \Large \int\limits\limits_{}^{}8tanxsec^3xdx\] tried u = sec x?

Answer 7

First though write it as, you'll see why when you find u and du\[\Large \int\limits\limits\limits_{}^{}8tanxsecx *\sec^2 xdx\]

Answer 8

ohh ok so \[\int\limits_{}^{}8u^2du\]

then

Answer 9

You're on the right track, but you're supposed to say
let u=sec(x), du=sec(x)tan(x)dx,
to help you remember what you did when you back-substitute.
Now all you have to do is to:
integrate and back-substitute!

Answer 10

yes, my bad. so integrating that:

\[\frac{ 8u^{3} }{ 3 }+ C\]

back-substitute:

\[\frac{ 8\sec^3x }{ 3 }\]+ C

Answer 11

\[\sqrt{}\]

Answer 12

yay thanks guys