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

解

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

- (- (- 0.51117 \dots + sin (1) (Ci (1) - Ci (0)) - cos^{2} (1) + cos (1) - sin^{2} (1)))

解答ステップ

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

u置換積分法を適用する

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

拡張

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

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

\int_{0}^{1} cos (1) sin (u) d u = - cos^{2} (1) + cos (1)

\int_{0}^{1} sin (1) cos (u) d u = sin^{2} (1)

= - (- (- 0.51117 \dots + sin (1) (Ci (1) - Ci (0)) - cos^{2} (1) + cos (1) - sin^{2} (1)))