证明∫(0,π/2)sin^m x cos^m x dx=1/2^m∫(0,π/2)cos^m xdx
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证明∫(0,π/2)sin^m x cos^m x dx=1/2^m∫(0,π/2)cos^m xdx
证明∫(0,π/2)sin^m x cos^m x dx=1/2^m∫(0,π/2)cos^m xdx
证明∫(0,π/2)sin^m x cos^m x dx=1/2^m∫(0,π/2)cos^m xdx
由2sin(x)cos(x) = sin(2x) => ( 2sin(x)cos(x) )^m = sin^m(2x)
左边=> ∫(0,π/2)sin^m x cos^m x dx = (1/2)^(m + 1)∫(0,π/2)sin^m (2x) d(2x)
= (1/2)^(m + 1)∫(0,π)sin^m (x) d(x)
由对称性在(0,π),sin^m(x)的积分相当于两倍在(0,π/2)的积分
原式 = (1/2)^(m)∫(0,π/2)sin^m (x) d(x) = -(1/2)^(m)∫(0,π/2)cos^m (π/2 - x) d(π/2 - x)
= -(1/2)^(m)∫(π/2,0)cos^m (x) d(x)=1/2^m∫(0,π/2)cos^m xdx
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