推 young11539 :謝謝 06/05 18:17
pi x pi pi x
∫x f(sinx) dx = x∫ f(sin t)dt | - ∫ ∫f(sint)dt dx
0 0 0 0 0
pi pi x
= pi*∫ f(sin t)dt - ∫ ∫f(sint) dt dx
0 0 0
pi pi t
= pi*∫ f(sin x)dx - ∫ ∫ f(sint) dx dt
0 0 0
pi pi
= pi*∫ f(sin x)dx - ∫ t f(sin t)dt
0 0
pi pi
=pi*∫ f(sinx)dx - ∫ xf(sinx)dx
0 0
pi pi
=> 2∫xf(sinx)dx = pi∫f(sinx)dx
0 0
pi pi
=> ∫xf(sinx)dx = pi/2 *∫f(sinx)dx
0 0
以上
※ 引述《young11539 (〝☆小小霈★”)》之銘言：
: if f(x) conti. on [0,1]
: then show that
: ∫xf(sinx)dx = (π/2)∫f(sinx)dx
: 積分範圍 [0,π]
: 感激不盡
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