证明∫(0,π/2)sin^m x dx=∫(0,π/2)cos^m x dx

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证明∫(0,π/2)sin^m x dx=∫(0,π/2)cos^m x dx
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证明∫(0,π/2)sin^m x dx=∫(0,π/2)cos^m x dx
证明∫(0,π/2)sin^m x dx=∫(0,π/2)cos^m x dx

证明∫(0,π/2)sin^m x dx=∫(0,π/2)cos^m x dx
令x = π/2 - u => dx = - du
∫(0~π/2) sin^m(x) dx
= ∫(π/2~0) [sin(π/2 - u)]^m (- du)
= -∫(π/2~0) [cos(u)]^m du
= ∫(0~π/2) cos^m(u) du
= ∫(0~π/2) cos^m(x) dx,得证