# Calc

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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/u - 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!

• Calc - ,

Break it into two integrals, integral (sec^2(x)tan(x))dx and integral(-tan(x))dx. Both can be solved using u substitution.

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