Recent content by Cali920

Thanks for all the help!

Is there anything left?

So then = u2/2 + C + ∫ du/u = u2/2 + ln (|u|) + C = u2/2 + ln (|cos x|) + C? = (sec x)2/2 + ln (|cos x|) + C?

u = cos x du = - sin x dx → -du = sin x dx = u2/2 + C - ∫ -du/u = u2/2 + C - (- [log (|u|) + C])?

wait I think there is a sign mishap somewhere if we put the integral sign back in...oh no.

Thank you! Is that a rule? u = cos x du = - sin x dx → - du = sin x dx = u2/2 + C + ∫ du/u = u2/2 + C + log (|u|) + C then plug back in u? = u2/2 + C + log (|cos x|) + C?

That's okay! Thank you! & I forgot to put the sign in there! THANK YOU Mark44! u = cos x du = - sin x dx → - du = sin x dx so then u2/2 + C + ∫ du/u Okay, but I'm not sure where to go from here or how to integrate du/u?

I made it positive I thought? because it was - - du/u?

Just didn't copy and paste right. But thanks again! I really appreciate it. =∫tan x (sec2x - 1) dx ∫(tan x sec2x - tan x) dx ∫tan x sec2x dx - ∫tan x dx u = sec x du = sec x tan x dx ∫tan x sec x sec x - ∫tan x dx ∫udu - ∫tan x dx ? = u2/2 + C -∫ (sin x / cos x) dx u =...

whoops! THANK YOU! =∫tan x (sec2x - 1) dx ∫(tan x sec2x - tan x) dx ∫tan x sec2x dx - ∫tan x dx u = sec x du = sec x tan x dx ∫tan x sec x sec x - ∫tan x dx ∫udu - ∫tan x dx ? = 1/2 u^2 + C -∫ (sin x / cos x) dx? so u = cos x?

I think I started off doing that originally then got stuck which is why I tried a different u substitution but ->(from my first post): =∫tan x (sec2x - 1) dx ∫(tan x (sec2x - tan x) dx ∫tan x sec2x dx - ∫tan x dx u = sec x du = sec x tan x dx ∫tan x sec x sec x - ∫tan x dx ∫udu - ∫tan x dx ?

Oh okay. So then what can I do if I can't do ∫ tan x dx * ∫ tan2x dx?

AH! I didn't even realize I did that. = ∫ tan x * ∫ tan2x = ∫ tan x (sec2x - 1) dx = ∫ (tan x sec2x) dx - ∫ (tan x) dx u = tan x du = sec2x dx = ∫ u * du - ∫ u Is this correct so far?Or can you do two u substitutions?: AH! I didn't even realize I did that. = ∫ tan x * ∫ tan2x = ∫ tan x...

so ∫ du * u - (sin x /cos x)? What do I do with the du * u? I'm confused... Thanks for the response! I appreciate it!

Homework Statement ∫tan3x dx 2. The attempt at a solution ∫tan x + ∫tan2x ∫tan x (sec2x - 1) dx ∫(tan x (sec2x - tan x) dx ∫tan x sec2x dx - ∫tan x dx u = sec x du = sec x tan x dx ∫tan x sec x sec x - ∫tan x dx Now I'm stuck.. ∫ du * u - ∫ tan x dx ?