求积分∫sint^2 dt怎么做？是t本身平方 不是sin平方

Question

如题所述

Answer 1

∫sint^2 dt
=tsint^2-∫tdsint^2
=tsint^2-2∫t^2*cost^2dt
=tsint^2-2∫t^2(1-sint^2)dt
=tsint^2-2∫t^2dt+2∫t^2sint^dt
=tsint^2-2t^3/3-∫td(cost^2)
=tsint^2-2t^3/3-tcost^2+∫cost^2dt
=tsint^2-2t^3/3-tcost^2+∫(1-sint^2)dt
=tsint^2-2t^3/3-tcost^2+∫dt-∫sint^2dt
=tsint^2-2t^3/3-tcost^2+t-∫sint^2dt

所以：
∫sint^2dt=(1/2)tsint^2-2t^3/6-(1/2)tcost^2+t/2.

Answer 2

∫sin(t^2)dt=∫(t)'sin(t^2)dt=tsint-2∫(t^2)cos(t^2)dt
And you can calculate.本回答被提问者采纳

Answer 3

怎么都做的这么烦，把sint^2=1-cost^2 先把1 提出去，剩下的就简单了。好好看看书吧 ，这是很简单的积分

Answer 4

S(x)+C