@[vitual_bot](/user/724676) 您是要对一切 $r$ 求积分吗 /yiw
by Rubidium_Chloride @ 2022-06-10 08:34:02
@[Reality_Creator](/user/278629) 差不多
by Iwara_qwq @ 2022-06-10 08:38:08
应该就是三角换元求不定积分
by Feng_Jing @ 2022-06-10 08:43:00
我思考一下，假如 $r$ 不变，$\sqrt{r^2-t^2}=(r^2-t^2)^{\frac{1}{2}}$ 然后把 $r^2-t^2$ 换个元就能做了吧
by Rubidium_Chloride @ 2022-06-10 08:43:59
@[Reality_Creator](/user/278629) 但是怎么换元
by Iwara_qwq @ 2022-06-10 08:51:56
@[vitual_bot](/user/724676) 等下我私信发你
by Feng_Jing @ 2022-06-10 08:52:43
咋自学积分啊，求教/kel
by fjy666 @ 2022-06-10 08:55:39
@[vitual_bot](/user/724676) $0.5r^2arcsin\frac{x}{r}+0.5x\sqrt{r^2-x^2}$
by Trinitrotoluene @ 2022-06-10 09:03:46
@[Trinitrotoluene](/user/141555) 一样的
by Feng_Jing @ 2022-06-10 09:05:27
@[FengJingFJ2022](/user/576077) 确实是用三角换元做的
by Trinitrotoluene @ 2022-06-10 09:07:13