integral from 0 to 1 of (sin(x))/(1-x)

解

\int_{0}^{1} \frac{sin ( x )}{1 - x} d x

- (- (- 0.51117 \dots + sin (1) (Ci (1) - Ci (0))))

解答ステップ

\int_{0}^{1} \frac{sin ( x )}{1 - x} d x

u置換積分法を適用する

= \int_{1}^{0} - \frac{sin ( - u + 1 )}{u} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{1} - \frac{sin ( - u + 1 )}{u} d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - (- \int_{0}^{1} \frac{sin ( - u + 1 )}{u} d u)

三角関数の公式を使用して書き換える

= - (- \int_{0}^{1} \frac{- cos ( 1 ) sin ( u ) + cos ( u ) sin ( 1 )}{u} d u)

拡張

\frac{- cos ( 1 ) sin ( u ) + cos ( u ) sin ( 1 )}{u} : - \frac{cos ( 1 ) sin ( u )}{u} + \frac{sin ( 1 ) cos ( u )}{u}

= - (- \int_{0}^{1} - \frac{cos ( 1 ) sin ( u )}{u} + \frac{sin ( 1 ) cos ( u )}{u} d u)

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - (- (- \int_{0}^{1} \frac{cos ( 1 ) sin ( u )}{u} d u + \int_{0}^{1} \frac{sin ( 1 ) cos ( u )}{u} d u))

\int_{0}^{1} \frac{cos ( 1 ) sin ( u )}{u} d u = 0.51117 \dots

\int_{0}^{1} \frac{sin ( 1 ) cos ( u )}{u} d u = sin (1) (Ci (1) - Ci (0))

= - (- (- 0.51117 \dots + sin (1) (Ci (1) - Ci (0))))