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

解

\int_{0}^{π} sin^{2} (x) cos (n x) d x

0

解答ステップ

\int_{0}^{π} sin^{2} (x) cos (n x) d x

三角関数の公式を使用して書き換える

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

部分積分を適用する

= [\frac{1}{n} (sin (n x) sin^{2} (x) - n \cdot \int \frac{1}{n} sin (n x) sin (2 x) d x)]_{0}^{π}

\int \frac{1}{n} sin (n x) sin (2 x) d x = \frac{1}{2 n} (- \frac{1}{n + 2} sin (n x + 2 x) + \frac{1}{n - 2} sin (n x - 2 x))

= [\frac{1}{n} (sin (n x) sin^{2} (x) - n \frac{1}{2 n} (- \frac{1}{n + 2} sin (n x + 2 x) + \frac{1}{n - 2} sin (n x - 2 x)))]_{0}^{π}

簡素化

[\frac{1}{n} (sin (n x) sin^{2} (x) - n \frac{1}{2 n} (- \frac{1}{n + 2} sin (n x + 2 x) + \frac{1}{n - 2} sin (n x - 2 x)))]_{0}^{π} : [\frac{1}{n} (sin (n x) sin^{2} (x) + \frac{1}{2 ( n + 2 )} sin (n x + 2 x) - \frac{1}{2 ( n - 2 )} sin (n x - 2 x))]_{0}^{π}

= [\frac{1}{n} (sin (n x) sin^{2} (x) + \frac{1}{2 ( n + 2 )} sin (n x + 2 x) - \frac{1}{2 ( n - 2 )} sin (n x - 2 x))]_{0}^{π}

限度を計算する:

0

= 0