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人気のある微分積分 >

integral from 0 to pi/2 of cos^9(x)

解

\int_{0}^{\frac{π}{2}} cos^{9} (x) d x

解

\frac{128}{315}

+1

十進法表記

0.40634 \dots

解答ステップ

\int_{0}^{\frac{π}{2}} cos^{9} (x) d x

三角関数の公式を使用して書き換える

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

u置換積分法を適用する

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

拡張

(1 - u^{2})^{4} : 1 - 4 u^{2} + 6 u^{4} - 4 u^{6} + u^{8}

= \int_{0}^{1} 1 - 4 u^{2} + 6 u^{4} - 4 u^{6} + u^{8} 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_{0}^{1} 1 d u - \int_{0}^{1} 4 u^{2} d u + \int_{0}^{1} 6 u^{4} d u - \int_{0}^{1} 4 u^{6} d u + \int_{0}^{1} u^{8} d u

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

\int_{0}^{1} 4 u^{2} d u = \frac{4}{3}

\int_{0}^{1} 6 u^{4} d u = \frac{6}{5}

\int_{0}^{1} 4 u^{6} d u = \frac{4}{7}

\int_{0}^{1} u^{8} d u = \frac{1}{9}

= 1 - \frac{4}{3} + \frac{6}{5} - \frac{4}{7} + \frac{1}{9}

簡素化

1 - \frac{4}{3} + \frac{6}{5} - \frac{4}{7} + \frac{1}{9} : \frac{128}{315}

= \frac{128}{315}

グラフ

人気の例

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