1. ## trig substitution

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.

2. $\int {\tan ^3 x~dx} = \int {\tan x(\sec ^2 x - 1)~dx} = \int {\tan x\sec ^2 x~dx} - \int {\tan x~dx}$

... everything you want... you got it...

3. Originally Posted by Krizalid
$\int {\tan ^3 x~dx} = \int {\tan x(\sec ^2 x - 1)~dx} = \int {\tan x\sec ^2 x~dx} - \int {\tan x~dx}$

... everything you want... you got it...
I'm confused because the answer in the book is 1/2tan^2(x) + ln(cos(x)) + C
I understand where the 1/2tan^2(x) but where does the ln(cos(x)) come from?

4. Curious...

$(\tan^2x)'=(\sec^2x)'$

Well $\int {\tan x~dx} = \int {\frac{{\sin x}}
{{\cos x}}~dx}$

That's easy, don't ya?