∫(0→π/2) [(sint)^4-(sint)^6] dt
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∫(0→π/2) [(sint)^4-(sint)^6] dt
这里用一个公式会简单些:∫ [0--->π/2] f(sinx)dx=∫ [0--->π/2] f(cosx)dx
∫[0→π/2] (sin⁴t-sin⁶t) dt
=∫[0→π/2] sin⁴t(1-sin²t) dt
=∫[0→π/2] sin⁴tcos²t dt
=1/2( ∫[0→π/2] sin⁴tcos²t dt+∫[0→π/2] sin²tcos⁴t dt )
=1/2 ∫[0→π/2] sin²tcos²t(sin²t+cos²t) dt
=1/2 ∫[0→π/2] sin²tcos²tdt
=1/8 ∫[0→π/2] sin²2tdt
=1/8 ∫[0→π/2] (1-cos4t)dt
=1/8(t-1/4sin4t) [0→π/2]
=π/16
∫[0→π/2] (sin⁴t-sin⁶t) dt
=∫[0→π/2] sin⁴t(1-sin²t) dt
=∫[0→π/2] sin⁴tcos²t dt
=1/2( ∫[0→π/2] sin⁴tcos²t dt+∫[0→π/2] sin²tcos⁴t dt )
=1/2 ∫[0→π/2] sin²tcos²t(sin²t+cos²t) dt
=1/2 ∫[0→π/2] sin²tcos²tdt
=1/8 ∫[0→π/2] sin²2tdt
=1/8 ∫[0→π/2] (1-cos4t)dt
=1/8(t-1/4sin4t) [0→π/2]
=π/16
∫sint/(cost+sint)dt
∫cost/(sint^2) dt =∫dsint/sint^2 =-1/sint + C
求定积分:∫π0(sint+cost)dt=
d/dx∫(上1下0)sint^2dt
limx→0[∫(0→x)cost^2dt]/[∫(0→x)(sint)/tdt]
极限x→0,求lim(∫(上x下0)sint^3dt)/x^4
∫dt/(1+sint+cost)
x→0时∫(上限x,下限-x)sint+sint^2dt 与ax^k 等价无穷小 求a与k
[(sint)^4-(sint)^6]从0 到π/2的积分是多少?[1-3cost+3(cost)^2-(cost)^3
d/dx∫(sint/t)dt上限π下限x
d[∫f(sint)dt]/dx,上限x,下限0
∫sint^2 cost^2 dt=?怎么算啊