# trig substitution

• Jul 26th 2007, 06:02 PM
davecs77
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.
• Jul 26th 2007, 06:09 PM
Krizalid
$\displaystyle \int {\tan ^3 x~dx} = \int {\tan x(\sec ^2 x - 1)~dx} = \int {\tan x\sec ^2 x~dx} - \int {\tan x~dx}$

... everything you want... you got it...
• Jul 26th 2007, 06:15 PM
davecs77
Quote:

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

... 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?
• Jul 26th 2007, 06:23 PM
Krizalid
Curious...

$\displaystyle (\tan^2x)'=(\sec^2x)'$ :D

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

That's easy, don't ya?