請幫忙計算不定積分∫xsin²xdx
- 2022-09-22
解
∫xsin²xdx
=(1/2)∫x(1-cos2x)dx
=(1/4)x²-(1/2)∫xcos2xdx
=(1/4)x²-(1/4)∫xdsin2x
=(1/4)x²-(1/4)xsin2x+(1/4)∫sin2xdx
=(1/4)(x²-xsin2x)-(1/8)cos2x+C
∫xsin²x dx
=∫x*(1-cos2x)/2 dx,利用三角函式恆等式cos2x=1-2sin²x
=(1/2)∫x dx - (1/2)∫xcos2x dx
=(1/2)(x²/2) - (1/2)(1/2)∫xcos2x d(2x),湊微分
=x²/4 - (1/4)∫x d(sin2x),湊微分
=x²/4 - (1/4)[xsin2x - ∫sin2x dx],分部積分法
=x²/4 - (1/4)[xsin2x - (1/2)∫sin2x d(2x)],湊微分
=x²/4 - (1/4)[xsin2x - (1/2)(-cos2x)] + c
=x²/4 - (1/4)xsin2x - (1/8)cos2x + c,注意頭一項的是x²而不是x