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人気のある微分積分 >

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

解

\int_{0}^{π} sin^{7} (x) d x

解

\frac{32}{35}

+1

十進法表記

0.91428 \dots

解答ステップ

\int_{0}^{π} sin^{7} (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))^{3} sin (x) d x

u置換積分法を適用する

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

拡張

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

= \int_{1}^{- 1} - 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

= - \int_{- 1}^{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

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

\int_{- 1}^{1} 1 d u = 2

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

\int_{- 1}^{1} 3 u^{4} d u = \frac{6}{5}

\int_{- 1}^{1} u^{6} d u = \frac{2}{7}

= - (- 2 + 2 - \frac{6}{5} + \frac{2}{7})

簡素化

- (- 2 + 2 - \frac{6}{5} + \frac{2}{7}) : \frac{32}{35}

= \frac{32}{35}

グラフ

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