Question

如何用分部积分法求原函数f(x)= x的导数？

Answer 1

令A=∫ (sinx)^4 dx，B=∫ (cosx)^4 dx
A+B=∫ (sinx)^4+(cosx)^4 dx
=∫嫌枝 [(sinx)^2+(cosx)^2]^2-2(sinxcosx)^2 dx
=∫ 1-(1/2)*(sin2x)^2 dx
=∫ 1-(1/4)*(1-cos4x) dx
=∫ 3/4+(1/4)*cos4x dx
=3x/4+(1/16)*sin4x+C1
A-B=∫ (sinx)^4-(cosx)^4 dx
=∫ [(sinx)^2+(cosx)^2][(sinx)^2-(cosx)^2] dx
=∫ (sinx)^2-(cosx)^2 dx
=(1/2)*∫ (1-cos2x)-(1+cos2x) dx
=∫ -cos2x dx
=(-1/2)*sin2x+C2
所以原式=A=(1/2)*[(A+B)+(A-B)]
=(1/2)*[3x/桐者举4+(1/16)*sin4x-(1/2)*sin2x]+C
=3x/8+(1/32)*sin4x-(1/4)*sin2x+C，其中局碧C是任意常数