求∫(arcsinx)^2dx=?
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求∫(arcsinx)^2dx=?
令arcsinx=t x=sint
∫(arcsinx)^2dx
=∫t^2costdt
=∫t^2dsint
=t^2sint-2∫tsintdt
=t^2sint+2∫tdcost
=t^2sint+2(tcost-∫costdt)
=t^2sint+2(tcost-sint)
∫(arcsinx)^2dx=x(arcsinx)^2+2(arcsinxcos(arcsinx)-x)
令x=sint,t∈(-Pi2,Pi/2]
则arcsinx=arcsin(sint)=t
∫(arcsinx)^2dx
=∫t^2dsint
=∫t^2(sint)'dt
=t^2*sint-∫2tsintdt
=t^2*sint+2∫t(cost)'dt
=t^2*sint+2tcost-2∫costdt
=t^2*sint+2t...
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令x=sint,t∈(-Pi2,Pi/2]
则arcsinx=arcsin(sint)=t
∫(arcsinx)^2dx
=∫t^2dsint
=∫t^2(sint)'dt
=t^2*sint-∫2tsintdt
=t^2*sint+2∫t(cost)'dt
=t^2*sint+2tcost-2∫costdt
=t^2*sint+2tcost-2∫dcost
=t^2*sint+2tcost-2cost
中间用了两次分部积分法
然后再代回去
∫(arcsinx)^2dx=
=(arcsinx)^2*x+2arcsinx*(1-x^2)^0.5-2*(1-x^2)^0.5
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