integral from 0 to pi/2 of 1 (sin^7(x))

解

\int_{0}^{\frac{π}{2}} \frac{d}{1} (sin^{7} (x)) d x

\frac{16}{35} d

解答ステップ

\int_{0}^{\frac{π}{2}} \frac{d}{1} sin^{7} (x) d x

定数を除く:

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

= d \cdot \int_{0}^{\frac{π}{2}} sin^{7} (x) d x

三角関数の公式を使用して書き換える

= d \cdot \int_{0}^{\frac{π}{2}} (1 - cos^{2} (x))^{3} sin (x) d x

u置換積分法を適用する

= d \cdot \int_{1}^{0} - (1 - u^{2})^{3} d u

拡張

- (1 - u^{2})^{3} : - 1 + 3 u^{2} - 3 u^{4} + u^{6}

= d \cdot \int_{1}^{0} - 1 + 3 u^{2} - 3 u^{4} + u^{6} d u

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

= d (- \int_{0}^{1} - 1 + 3 u^{2} - 3 u^{4} + u^{6} 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

= d (- (- \int_{0}^{1} 1 d u + \int_{0}^{1} 3 u^{2} d u - \int_{0}^{1} 3 u^{4} d u + \int_{0}^{1} u^{6} d u))

\int_{0}^{1} 1 d u = 1

\int_{0}^{1} 3 u^{2} d u = 1

\int_{0}^{1} 3 u^{4} d u = \frac{3}{5}

\int_{0}^{1} u^{6} d u = \frac{1}{7}

= d (- (- 1 + 1 - \frac{3}{5} + \frac{1}{7}))

簡素化

d (- (- 1 + 1 - \frac{3}{5} + \frac{1}{7})) : \frac{16}{35} d

= \frac{16}{35} d