求不定积分?∫ x sin(x/2) dx

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求不定积分?∫ x sin(x/2) dx

求不定积分?∫ x sin(x/2) dx
求不定积分?∫ x sin(x/2) dx

求不定积分?∫ x sin(x/2) dx
∫ x sin(x/2) dx
=4∫(x/2)sin(x/2)d(x/2) 令t=x/2
=4∫t sint dt
= -4∫td(cost)
= -4(t cost -∫cost dt)
=4sint - 4t cost+C
=4sin(x/2)-2xcos(x/2)+C

∫ x sin(x/2) dx
=2∫ x sin(x/2) d(x/2)
=-2∫ x dcos(x/2)
=-2(x*cos(x/2)-∫cos (x/2 ) dx)
=-2x*cos(x/2)+4sin(x/2)+c c为常数
不定积分记得不太清了 算个大概吧

以后做这样关于三角函数的积分尽量先转化三角函数