integral from 0 to 2pi of sin(x)cos(nx)

解

\int_{0}^{2 π} sin (x) cos (n x) d x

\frac{n cos ( 2 π + 2 πn ) - cos ( 2 π + 2 πn ) + 2 - n cos ( 2 π - 2 πn ) - cos ( 2 π - 2 πn )}{2 ( n + 1 ) ( - n + 1 )}

解答ステップ

\int_{0}^{2 π} sin (x) cos (n x) d x

次の恒等を使用する:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

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

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

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

= \frac{1}{2} (\frac{- cos ( 2 π + 2 πn ) + 1}{1 + n} + \frac{- cos ( 2 π - 2 πn ) + 1}{1 - n})

簡素化

\frac{1}{2} (\frac{- cos ( 2 π + 2 πn ) + 1}{1 + n} + \frac{- cos ( 2 π - 2 πn ) + 1}{1 - n}) : \frac{n cos ( 2 π + 2 πn ) - cos ( 2 π + 2 πn ) + 2 - n cos ( 2 π - 2 πn ) - cos ( 2 π - 2 πn )}{2 ( n + 1 ) ( - n + 1 )}

= \frac{n cos ( 2 π + 2 πn ) - cos ( 2 π + 2 πn ) + 2 - n cos ( 2 π - 2 πn ) - cos ( 2 π - 2 πn )}{2 ( n + 1 ) ( - n + 1 )}