# Integral from -pi/3 to 0 of 3sin^3(x)dx ?

1. Feb 19, 2013

### Lo.Lee.Ta.

∫from $-\pi$/3 to 0 of 3sec3xdx

Integration by parts...

3∫sec2x*secxdx

dv= 3sec2x
v= 3tanx

u= secx
du= secxtanx

Formula: uv - ∫vdu

secx*3tanx - ∫3tanx*secxtanxdx

Integration by parts again...

u= 3tanx
du= 3sec2x

dv= sextanx
v= secx

3tanx*secx - ∫secx*3sec2x

Altogether:

∫3sec3x = 3secxtanx - 3secxtanx - ∫3sec3x

∫2*3sec3x = 0 !!!

It seems that the two 3secxtanx's would cancel! :/

∫6sec3x = 0

This is like the same thing we started out with! ugh, man. I don't really know what to do now.
I know this is a definite integral, so we should substitute -pi/3 and 0 in the end, but we still have an integral! #=_=

...This definitely does not seem right! :(
But I don't really see what's wrong with my work.

Last edited by a moderator: Feb 19, 2013
2. Feb 19, 2013

### Staff: Mentor

That's not the way to go. Your integral above can be written as
$\int tan^2(x) sec(x) dx$
$= \int (sec^2(x) - 1) sec(x) dx$
This will give you another sec3 integral and one that's somewhat easier. You should be able to solve algebraically for the integral you're looking for.
I am not including the factor of 3 from the first integral. It would have been better to bring it out of the integral right away and just concentrate on ∫sec3(x) dx.