这是高中题目么，要用到分部积分的知识吧
∫(sinx)^5dx=-∫(sin^4xdcosx)=-(sinx)^4cosx+4∫(sinx)^3(cosx)^2dx
=-(sinx)^4cosx+4∫(sinx)^3(1-sinx^2)dx=-(sinx)^4cosx+4∫(sinx)^3dx-4∫(sinx)^5dx
所以5∫(sinx)^5dx=-(sinx)^4cos...
这是高中题目么，要用到分部积分的知识吧
∫(sinx)^5dx=-∫(sin^4xdcosx)=-(sinx)^4cosx+4∫(sinx)^3(cosx)^2dx
=-(sinx)^4cosx+4∫(sinx)^3(1-sinx^2)dx=-(sinx)^4cosx+4∫(sinx)^3dx-4∫(sinx)^5dx
所以5∫(sinx)^5dx=-(sinx)^4cos...