Asked by UCI Student
Hello im trying to integrate tan^3 dx
i have solved out the whole thing but it doesnt match up with the solution..
this is what i did:
first i broke it up into:
integral tan^2x (tanx) dx
integral (sec^2x-1)(tanx) dx
then i did a u substitution
u = secx
du = secxtanx dx (dx = du/secxtanx)
so now i have..
integral (u^2 - 1)*tanx* du/secxtanx
(then the tanx's cancel and then i have a secx with which i re-subsitute u for)
so now i have:
integral (u^2-1)/u
and break it up...
integral u^2/2 - integral 1/u
= integral u - integral 1/u
=u^2/2 - lnabs(u) + c
(and then plug u back in)
(sec^2x)/2 - lnabs(sec^2x) + c
...but it says this is wrong because the anser is actually tan^2x/2 - lnabs(sec^2x) + c
ive done this problem numerous times and i just cant figure out what im doing wrong, any help would be amazing thank you so much!
i have solved out the whole thing but it doesnt match up with the solution..
this is what i did:
first i broke it up into:
integral tan^2x (tanx) dx
integral (sec^2x-1)(tanx) dx
then i did a u substitution
u = secx
du = secxtanx dx (dx = du/secxtanx)
so now i have..
integral (u^2 - 1)*tanx* du/secxtanx
(then the tanx's cancel and then i have a secx with which i re-subsitute u for)
so now i have:
integral (u^2-1)/u
and break it up...
integral u^2/2 - integral 1/u
= integral u - integral 1/u
=u^2/2 - lnabs(u) + c
(and then plug u back in)
(sec^2x)/2 - lnabs(sec^2x) + c
...but it says this is wrong because the anser is actually tan^2x/2 - lnabs(sec^2x) + c
ive done this problem numerous times and i just cant figure out what im doing wrong, any help would be amazing thank you so much!
Answers
Answered by
UCI Student
(on the 17th line it shud be u^2/u)
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