Question

求sinx的m次方与 cosx的n次方的乘积的不定积分,求sin(ax)*cos(bx)的不定积分

Answer 1

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