求微分∫(sin√x)^2dx

Question

如题所述

Answer 1

应该是 ： 求不定积分 ∫(sin√x)^2dx
令 t=√x， 则
I = ∫(sin√x)^2dx = ∫2t(sint)^2dt = ∫t(1-cos2t)dt
= t^2/2 - (1/2)∫tdsin2t
= t^2/2 - (1/2)tsin2t + (1/2)∫sin2tdt
= t^2/2 - (1/2)tsin2t - (1/4)cos2t +C
= x/2 - (1/2)√xsin(2√x) - (1/4)cos(2√x)+C

Answer 2

d ∫ sin(√x)^2 dx / dx = sin(√x)^2