求不定积分,∫xsin눀xdx.

2024-11-08 17:57:52
推荐回答(1个)
回答(1):

[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(2x)/8 +C