integral from 0 to 2pi of sin(x)sin(x+1)

解

\int_{0}^{2 π} sin (x) sin (x + 1) d x

π cos (1) + \frac{- sin ( 4 π + 1 ) + sin ( 1 )}{4}

十進法表記

1.69740 \dots

解答ステップ

\int_{0}^{2 π} sin (x) sin (x + 1) d x

三角関数の公式を使用して書き換える

= \int_{0}^{2 π} \frac{1}{2} (- cos (2 x + 1) + cos (1)) d x

定数を除く:

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

= \frac{1}{2} \cdot \int_{0}^{2 π} - cos (2 x + 1) + cos (1) d x

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

= \frac{1}{2} (- \int_{0}^{2 π} cos (2 x + 1) d x + \int_{0}^{2 π} cos (1) d x)

\int_{0}^{2 π} cos (2 x + 1) d x = \frac{1}{2} (sin (4 π + 1) - sin (1))

\int_{0}^{2 π} cos (1) d x = 2 π cos (1)

= \frac{1}{2} (- \frac{1}{2} (sin (4 π + 1) - sin (1)) + 2 π cos (1))

簡素化

\frac{1}{2} (- \frac{1}{2} (sin (4 π + 1) - sin (1)) + 2 π cos (1)) : π cos (1) + \frac{- sin ( 4 π + 1 ) + sin ( 1 )}{4}

= π cos (1) + \frac{- sin ( 4 π + 1 ) + sin ( 1 )}{4}