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人気のある微分積分 >

integral from 0 to pi/2 of cos^{2022}(x)sin(x)

解

\int_{0}^{\frac{π}{2}} cos^{2022} (x) sin (x) d x

解

\frac{1}{2023}

+1

十進法表記

0.00049 \dots

解答ステップ

\int_{0}^{\frac{π}{2}} cos^{2022} (x) sin (x) d x

u置換積分法を適用する

= \int_{1}^{0} - u^{2022} d u

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

= - \int_{0}^{1} - u^{2022} d u

定数を除く:

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

= - (- \int_{0}^{1} u^{2022} d u)

べき乗則を適用する

= - (- [\frac{u ^{2023}}{2023}]_{0}^{1})

簡素化

= [\frac{u ^{2023}}{2023}]_{0}^{1}

限度を計算する:

\frac{1}{2023}

= \frac{1}{2023}

グラフ

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