根据：sinAcosB=[sin(A+B)+sin(A-B)]/2
上式=积分[sin(mx+nx)+sin(mx-nx)]/2=
1/(2m+2n)+1/(2m-2n)-cos(m*pi+n*pi)/(2m+2n)-cos(m*pi-n*pi)/(2m-2n)
超一流数学高手、数学专家请进(Integration 0026)
根据：sinAcosB=[sin(A+B)+sin(A-B)]/2
上式=积分[sin(mx+nx)+sin(mx-nx)]/2=
1/(2m+2n)+1/(2m-2n)-cos(m*pi+n*pi)/(2m+2n)-cos(m*pi-n*pi)/(2m-2n)