来自贾志勇的问题
求不定积分,∫xsin²xdx.
求不定积分,∫xsin²xdx.
1回答
2020-02-02 07:05
求不定积分,∫xsin²xdx.
求不定积分,∫xsin²xdx.
[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)]/2dx=(1/2)∫[x-xcos(2x)]dx=(1/2)[x²/2-xsin(2x)/2-cos(2x)/4]+C=x²/4-xsin(2x)/4-cos(...