求不定积分,∫xsin²xdx.
问题描述:
求不定积分,∫xsin²xdx.
答
[x²/2-xsin(2x)/2-cos(2x)/4]'=x-sin(2x)/2-xcos(2x)+sin(2x)/2=x-xcos(2x)∫xsin²xdx=∫x[1-cos(2x)]/2 dx=(1/2)∫[x-xcos(2x)]dx=(1/2)[x²/2-xsin(2x)/2-cos(2x)/4]+C=x²/4- xsin(2x)/4- cos(...