integral from 0 to pi/4 of sin^5(x)

解

\int_{0}^{\frac{π}{4}} sin^{5} (x) d x

\frac{8}{15} - \frac{43}{60 2}

十進法表記

0.02657 \dots

解答ステップ

\int_{0}^{\frac{π}{4}} sin^{5} (x) d x

三角関数の公式を使用して書き換える

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

u置換積分法を適用する

= \int_{1}^{\frac{2}{2}} - 1 + 2 u^{2} - u^{4} d u

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

= - \int_{\frac{2}{2}}^{1} - 1 + 2 u^{2} - u^{4} 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_{\frac{2}{2}}^{1} 1 d u + \int_{\frac{2}{2}}^{1} 2 u^{2} d u - \int_{\frac{2}{2}}^{1} u^{4} d u)

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

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

\int_{\frac{2}{2}}^{1} u^{4} d u = \frac{1}{5} - \frac{1}{20 2}

= - (- (1 - \frac{1}{2}) + 2 (\frac{1}{3} - \frac{1}{6 2}) - (\frac{1}{5} - \frac{1}{20 2}))

簡素化

- (- (1 - \frac{1}{2}) + 2 (\frac{1}{3} - \frac{1}{6 2}) - (\frac{1}{5} - \frac{1}{20 2})) : \frac{8}{15} - \frac{43}{60 2}

= \frac{8}{15} - \frac{43}{60 2}