Question

\[\int sin^6 2x dx\]
what's an easy way to do this integral
instead of doing:
\[\int sin^n u du= -\frac 1 n sin^{n-1} ucosu + \frac{n-1}{n} \int sin^{n-2} u du +C\]

Answer 1

lol that is the easy way!

Answer 2

NO WAY.....what?

Answer 3

reduction formula or whatever
that is what you have to do

Answer 4

ok i will be quiet, maybe someone else has a snap way to do it
i doubt it
we can check the answer and see

Answer 5

okay...guess I'll have to do this twice then...
my first answer consisted of a
\[\int sin^4 2x dx\]

Answer 6

just as I was starting to like math.......sigh

Answer 7

I'm just handwaving here, and I too believe that reduction formula is the way to go. Nevertheless, making an attempt to devise something simpler. \[\LARGE sin(x) = \frac{e^{ix} - e^{-ix}}{2} \Rightarrow sin^{6}(2x) = (\frac{e^{2ix}-e^{-2ix}}{2})^{6}\]
Maybe try a binomial expansion of this and integrate them. The integration will be easy enough as its exponential terms. Its the getting the result back in terms of sin/cos that will be an issue.

Answer 8

hey buddy,
thanks for the offer, but what you have written above looks just a tid too complicated for me

Answer 9

I assumed that you would know complex numbers, \(i = \sqrt{-1}\) and the fact that \(e^{i\theta } = cos\theta + i.sin\theta \)

Answer 10

is that calc II material?

Answer 11

I'm not sure. If you don't know complex numbers, then I don't see any other method other than your first post or by repeated application of integration by parts

Answer 12

yeah, it's not too bad though, I completed one problem already

Answer 13

lol definitely not calc 2 material, but a good way to proceed if you want to turn difficult problems into what often comes down to the laws of exponents
they didn't put these reduction formulas on the back page of your text just to waste type

Answer 14

Lol...good to know