Ok, it seems it an orthogonal problem.
(%i7) integrate(sin(n*%pi*x)*sin(%pi*x),x,-1,1) ;
(%o7) 0
I tried below.
(%i2) declare(n,integer) ;
(%o2) done
(%i3) integrate(sin(n*%pi*x/a)*(sin(%pi*x/a))^3,x,-a,a) ;
(%o3) 0
Now, (sin(sin(%pi*x/a))^3 can be converted to (3*sin(%pi*x/a)-sin(3*%pi*x/a))/4 and if I do above
integration manually, it yields that the result is 0 for any n except 1 or 3. If n=1 result should be 3/4*a
and if n=3 result should be -a/4 but not zero for both cases. Is this a bug?
Maxima version: 5.17.1
Maxima build date: 20:10 3/10/2009
host type: powerpc-apple-darwin8.11.0
lisp-implementation-type: CMU Common Lisp
lisp-implementation-version: Stage 3 20071108T014921 (19D)
Ok, it seems it an orthogonal problem.
(%i7) integrate(sin(n*%pi*x)*sin(%pi*x),x,-1,1) ;
(%o7) 0
When n is not declared to be an integer the answer is:
(%i6) integrate(sin(n*%pi*x/a)*(sin(%pi*x/a))^3,x,-a,a);
(%o6) 12*a*sin(%pi*n)/(%pi*n^4-10*%pi*n^2+9*%pi)
This answer simplifies correctly to zero when n is an integer.
Closing this bug report as invalid.
Dieter Kaiser
Reopening bug.
Shouldn't maxima at least ask if n = 1 or something? When n = 1, the integrand is sin(%pi*x/a)^4, so the integral can't be zero.
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