Question

prove this reduction formula, using integration by parts

Answer 1

Sure. Do you know the integration by parts formula?

Answer 2

yes

Answer 3

Sorry, mu computer is super slow for some reason. But yeah, use the cos function as your dv and just use 1 as u. Hope that helps.

Answer 4

\[\int\limits_{}^{}\cos ^{n}x\ dx=x \cos ^{n}x+n \int\limits_{}^{}x \cos ^{n-1}x \sin x \ dx\]I'm feeling your pain...

Answer 5

Ok... I got it... but it's huge. Here are some steps and hints for you to try.
Take what I have above and use by parts again on the right side integral.\[u=\cos ^{n-1}x \ \ dv=x \sin x\]and you of course have to do x sin x by parts.

After you do this (carefully distributing n's and minus signs) you will have an enormous equation that you will have to turn your paper sideways for.

Now for some key substitutions...
\[\cos ^{n-1}x \cos x=\cos ^{n}x\]and\[\cos ^{n-2}x \cos x=\cos ^{n-1}x\]and\[\sin ^{2}x=1-\cos ^{2}x\]and the craziest substitution... take the answer from the first by parts solution, solve for the right hand integral and switch it out for its equivalent in the enormous equation.\[n \int\limits_{}^{}x \cos ^{n-1}x \sin x dx=\int\limits_{}^{}\cos ^{n}xdx-x \cos ^{n}x\]stir it well and cancel several terms until you clearly see all of the terms involving pure\[\int\limits_{}^{}\cos ^{n}x \ dx\]take all of those terms to the left hand side and simplify.

I am trying to write it neatly on paper (as the mess is on my white board). Let me know if you don't get it and I will scan my answer.

Answer 6

Thank you @EulerGroupie

Answer 7

You're welcome... this was a good one ;)

Answer 8

Ok... I'm proud... gotta show it off.