∫(0→π/2) [(sint)^4－(sint)^6] dt
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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
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