Question

integral of e^3x sec(e^3x) tan(e^3x)dx

Answer 1

u = sec(e^3x)
du =e^3x sec(e^3x) tan(e^3x dx?

\[\int\limits du\]?? = u + c = sec(e^3x) +c?

Answer 2

du=3e^3x sec(e^3x) tan(e^3x) dx
but other than that, yes

Answer 3

\[3 \int\limits du\]??

Answer 4

err actually...

Answer 5

\[1/3 \int\limits du\]

Answer 6

yep

Answer 7

I guess I forgot the chain rule oopsies :P

Answer 8

you did well to leave it at du
most would have tried to integrate u

Answer 9

\[1/3 \sec(e^{3x}) +c\]

Answer 10

correct :)

Answer 11

Yeah I was first thinking "Wait what about u?" Then I checked the front of the book to which it stated int du = u+c so I figured I'd go with that and make sure :D.

Answer 12

btw for the last question tan(arcsec(x/8)) I wrote out y = arcsec(x/8) then sec y = (8/x) then tan y = that big fraction? Is that correct how to write it out?

Answer 13

That's what the book says, so hopefully :P.

Answer 14

sure, that's how I wrote it out
with the picture next to it, and y labeled, it makes more senseI guess I would prefer to call the angle u instead of y, so I'd write\[\sec^{-1}u=\frac x8\implies u=\sec\frac x8\implies\tan u=\frac{\sqrt{x^2-8^2}}8\]