Let Im,n=∫(sinx)^m*(cosx)^ndx then Im,n=(sinx)^(m+1)*(cosx)^(n-1)- ∫(sinx)[(sinx)^m*(cosx)^(n-1)]'dx =(sinx)^(m+1)*(cosx)^(n-1)- ∫[m(sinx)^m*(cosx)^n-(n-1)(sinx)^(m+2)*(cosx)^(n-1)]dx =(sinx)^(m+1)*(cosx)^(n-1)-mIm,n+(n-1)Im+2,n-2 so (m+1)Im,n=(sinx)^(m+1)*(cosx)^(n-1)+(n-1)Im+2,n-2 用此递推公式求解 sin(ax)*cos(bx) =(1/2)*[sin(a+b)x+sin(a-b)x] so ∫sin(ax)*cos(bx)dx =-(1/2)*[cos(a+b)x/(a+b)+cos(a-b)x/(a-b)]+C