How would you do this
integral of tan^3(x)dx
I broke it down into integral of tan(x)(sec^2(x) - 1)dx
u = tan(x)
du = sec^2(x)
so i got integral of u(du) - integral of (u)
I am not so sure if this is correct so far, and I do not know where to go from here. Thanks for the help.
... everything you want... you got it...
I'm confused because the answer in the book is 1/2tan^2(x) + ln(cos(x)) + C
Originally Posted by Krizalid
I understand where the 1/2tan^2(x) but where does the ln(cos(x)) come from?
That's easy, don't ya?