Question

∫xsin∧4(x)dx= ？急求 谢谢啦！

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Answer 1

∫x(sinx)^4dx
=(1/4)∫x(1-cos2x)^2dx
=(1/4)∫[x-2xcos2x +x(cos2x)^2 ]dx
=(1/8)∫[2x-4xcos2x +x(1+cos4x) ]dx
=(1/8)∫[3x-4xcos2x +xcos4x ]dx
=(1/8)[(3/2)x^2-∫ 4xcos2xdx +∫xcos4x dx ]

consider
∫4xcos2xdx
=2∫xdsin2x
=2xsin2x - 2∫sin2x dx
=2xsin2x +cos2x

∫xcos4x dx
=(1/4)∫xdsin4x
=(1/4)xsin4x-(1/4)∫sin4xdx
=(1/4)xsin4x+(1/16)cos4x

∫x(sinx)^4dx
=(1/8)[(3/2)x^2-∫ 4xcos2xdx +∫xcos4x dx ]
=(1/8)[(3/2)x^2-2xsin2x -cos2x +(1/4)xsin4x+(1/16)cos4x ] + C