∫(0→π/2) [(sint)^4-(sint)^6] dt
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∫(0→π/2) [(sint)^4-(sint)^6] dt
∫(0→π/2) [(sint)^4-(sint)^6] dt
∫(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) [(sint)^4-(sint)^6]
∫(0→π/2) [(sint)^4-(sint)^6] dt
∫(上限π/2 下限0) [(sint)^4-(sint)^6] dt
证明∫[π/10,x]sint²dt+∫[π/2,x]1/sint²dt=0在(π/10,π/2)内有唯一实根
高数高手进,求具体积分过程 题为∫(t-sint)² sint dt题为定积分计算上限2π,下限0
[(sint)^4-(sint)^6]从0 到π/2的积分是多少?[1-3cost+3(cost)^2-(cost)^3]从0到2π的积分是多少?
limx→0[∫(0→x)cost^2dt]/[∫(0→x)(sint)/tdt]
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lim(x→0)x^2tantx /∫(x,0)t(t+sint)dt
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d/dx∫(上1下0)sint^2dt
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∫sint/(cost+sint)dt
∫[(sinx)^3-(sinx)^5]dx∫x^3(1-x^2)^1/2 dx设x=sint,(1-x^2)^1/2=costdx=cost dt原式∫x^3(1-x^2)^1/2 dx=∫(sint)^3 cont cost dt=∫(sint)^3 (cont)^2 dt这步之后.不确定=∫(sint)^3 [1-(sint)^2] dt=∫[(sint)^3-(sint)^5] dt= -1/4(cost)^4+1/6(
试求函数y=∫(0→x)sint dt 当x=π/4时的导数 就是这样的 微积分啊
如何直接看出0到pai/2定积分cost/(sint+cost)与sint/(sint+cost)相等?
求∫(e^t*sint)^2 dt
为什么∫(2sint+cost)+2(2cost-sint)/2sint+cost dt = (t+2ln|2sint+cost|)+C?