For prove

* Proof. * First it will help if we simplify the integral:

Then we use the method of substitution, letting

So we have,

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Stumbling Robot

A Fraction of a Dot
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Prove an identity for integrals of trig functions

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### Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):

For prove

* Proof. * First it will help if we simplify the integral:

Then we use the method of substitution, letting

So we have,

I was thinking of doing something different:

Note that 1/2 * integral of cos(x) from 0 to pi/2 = integral of cos(x) * sin(x) from 0 to pi/2 = 1/2 (a)

Now cos^m(x) / (2^m) – cos^m(x) * sin^m(x) = (cos(x) / 2 – cos(x) * sin(x)) * cos^(m-1) (x) * (1/(2^m-1) + sin(x)/(2^m-2) + sin^2(x)/(2^m-3) + … + sin^(m-1) (x)), where cos^(m-1) (x) and

(1/(2^m-1) + sin(x)/(2^m-2) + sin^2(x)/(2^m-3) + … + sin^(m-1) (x)) are not negative on the interval from 0 to pi/2.

If one tries to evaluate the integral of cos^m(x) / (2^m) – cos^m(x) * sin^m(x) then one can use the cauchy-mean value theorem and take out both cos^(m-1) (x) and

(1/(2^m-1) + sin(x)/(2^m-2) + sin^2(x)/(2^m-3) + … + sin^(m-1) (x)) of the integral, by having them evaluated at some point c, between 0 and pi/2:

integral of cos^m(x) / (2^m) – cos^m(x) * sin^m(x) from 0 to pi/2

= cos^(m-1) (c) * (1/(2^m-1) + sin(c)/(2^m-2) + sin^2(c)/(2^m-3) + … + sin^(m-1) (c)) * integral of cos(x) / 2 – cos(x) * sin(x) from 0 to pi/2. From (a), we know that since the integral of cos(x) from 0 to pi/2 = integral of cos(x) * sin(x) from 0 to pi/2 that this is equal to 0, and hence:

integral of cos^m(x) / (2^m) from 0 to pi/2 = integral of cos^m(x) * sin^m(x) from 0 to pi/2.

Beautiful!