| ■24270 / inTopicNo.2) |
Re[1]: 積分と証明
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□投稿者/ ゼロ 軍団(133回)-(2007/04/26(Thu) 09:38:47)
 | ∫[0〜π/2] sin(2n+1)x/sinx dx=∫[0〜π/2] [sin(2n-1)xcos2x+cos(2n-1)xsin2x]/sinx dx
= ∫[0〜π/2] [sin(2n-1)x/sinx-2sin(2n-1)xsinx+2cos(2n-1)xcosx] dx
=∫[0〜π/2]sin(2n-1)x/sinx dx+∫[0〜π/2]cos2nx dx
=∫[0〜π/2]sin(2n-1)x/sinx dx
あとは
∫[0〜π/2] sinx/sinx dx=π/2なので示せました。
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