That is similar to "Hard trigonometry integration" at sci.math, 22 Nov 2009 and the elementary solution given by Leon Aigret 22 Nov 2009 (based on the first solution given by Robert Israel, who used residue calculus):

f:= x -> cos(2*x)*cos(cos(3*x))^2 = 1/4*cos(2*x-2*cos(3*x))+1/4*cos(2*x+2*cos(3*x))+1/2*cos(2*x)

Then (*) f(x) + f(x + 1/3*Pi) + f(x + 2/3*Pi) = 0 using cos(t) + cos (t + 2/3*Pi) + cos (t + 4/3*Pi) = 0.

Now observe the function is symmetric w.r.t. Pi/2 and the integral thus is just 1/2 of taking the integral over 0 ... Pi.

Then split in 1/3 and 2/3. Transforming the integrals over the last two intervals to live over 0 .. Pi/3 and using (*) gives the assertion, since the over all new integrand simplifies to 0.

Wow, do you or Aigret have anything to say re (search for) Coxeter's challenge integral in http://www.tweedledum.com/rwg/idents.htm ? --rwg

That is similar to "Hard trigonometry integration" at sci.math, 22 Nov 2009 and the elementary solution given by Leon Aigret 22 Nov 2009 (based on the first solution given by Robert Israel, who used residue calculus):

f:= x -> cos(2*x)*cos(cos(3*x))^2 = 1/4*cos(2*x-2*cos(3*x))+1/4*cos(2*x+2*cos(3*x))+1/2*cos(2*x)

Then (*) f(x) + f(x + 1/3*Pi) + f(x + 2/3*Pi) = 0 using cos(t) + cos (t + 2/3*Pi) + cos (t + 4/3*Pi) = 0.

Now observe the function is symmetric w.r.t. Pi/2 and the integral thus is just 1/2 of taking the integral over 0 ... Pi.

Then split in 1/3 and 2/3. Transforming the integrals over the last two intervals to live over 0 .. Pi/3 and using (*) gives the assertion, since the over all new integrand simplifies to 0.

Wow, do you or Aigret have anything to say re (search for) Coxeter's challenge integral in http://www.tweedledum.com/rwg/idents.htm ? --rwg

Alex, thanks so much for the solution and the detailed info. :) Regarding Dan's question: This integral was derived from \int_{0}^{\pi} \cos(2x+2\sin(3x)) dx. After some tranformations, I thought that the crux was the integral I posted since other integrals involved could be solved with simple symmetry. The original integral: \int_0^{\pi} \sin^4 (x+\sin 3x) dx was from a recent high school competition in Thailand. From the info kindly gave by Alex, now I know there must be some Thai posted this to sci.math soon after the competition. Warut P.S. 8*(sin A)^4 = 3 - 4*cos 2A + cos 4A, so \int_0^{\pi} \sin^4 (x+\sin 3x) dx = 3\pi /8. On Sat, Nov 28, 2009 at 4:01 AM, Axel Vogt <mail@axelvogt.de> wrote:

That is similar to "Hard trigonometry integration" at sci.math, 22 Nov 2009 and the elementary solution given by Leon Aigret 22 Nov 2009 (based on the first solution given by Robert Israel, who used residue calculus):

f:= x -> cos(2*x)*cos(cos(3*x))^2 =  1/4*cos(2*x-2*cos(3*x))+1/4*cos(2*x+2*cos(3*x))+1/2*cos(2*x)

Then (*) f(x) + f(x + 1/3*Pi) + f(x + 2/3*Pi) = 0 using  cos(t) + cos (t + 2/3*Pi) + cos (t + 4/3*Pi) = 0.

Now observe the function is symmetric w.r.t. Pi/2 and the integral thus is just 1/2 of taking the integral over 0 ... Pi.

Then split in 1/3 and 2/3. Transforming the integrals over the last two intervals to live over 0 .. Pi/3 and using (*) gives the assertion, since the over all new integrand simplifies to 0.

On Sat, Nov 28, 2009 at 3:12 AM, Dan Asimov <dasimov@earthlink.net> wrote:

Warut wrote:

<< Numerically, I found that

\int_{0}^{\pi/2} \cos(2x)(\cos(\cos(3x)))^2 dx = 0,

but I couldn't prove it. So I post it here and hope some mathfunsters could shed some light on it

...

...

Interesting. The curve has a cute graph that makes this conjecture plausible, but not obvious.

I'm curious: What led you to calculate this particular integral numerically?

--Dan