# Integral Help - integral of (cos^6)(x)

• Oct 19th 2013, 03:53 AM
toga
Integral Help - integral of (cos^6)(x)
What is the integral cos6(x)? I just want to make sure I am correct.
The answer I got was [(3sin(4x) + 2sin(2x) + 5x)/16] + C
but I have no clue if I am correct and am not very confident in my integration skills.
If your answer is different, it'd be a great help if you could show me the process you took to get there.

Thanks!
• Oct 19th 2013, 04:43 AM
Re: Integral Help - integral of (cos^6)(x)
No that's not correct try using the reduction formula.
• Oct 19th 2013, 04:47 AM
Re: Integral Help - integral of (cos^6)(x)
hint:
$\displaystyle \int cos^{m}(x)dx=\frac{sin(x)cos^{m-1}(x)}{m}+\frac{m-1}{m}\int cos^{m-2}(x)dx$
• Oct 19th 2013, 04:54 AM
MINOANMAN
Re: Integral Help - integral of (cos^6)(x)
No I don't think it is correct because the derivative of [(3sin(4x) + 2sin(2x) + 5x)/16] + C
is not equal to cos6(x) . Moreover you didn't show to us how you did it...
Anyway use repeated integration by parts and you will get it...but be patient because the process is lengthy...

Good Luck
• Oct 19th 2013, 05:01 AM
toga
Re: Integral Help - integral of (cos^6)(x)
SMAD - I don't think the reduction formula is within the scope of my course, so if this question comes up in my exam I don't think I will get marks for process.
Do you know of another way to do it?

MINOANIMAN- Yeah, I thought so, My working took more than a page (possibly because of my relatively large handwriting) so I didn't post that sorry, but I broke up the cos^6(x) into (cos^2(x))^3 and substituted the half angle formula (cos^2(x) = 1/2(1+cos(2x)) and went from there, but I may have made some algebraic errors that I can't see... :\ I'm just wondering if there's a better way to break it up that will result in less possible errors.