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

解答

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

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

十进制

1.04009 \dots

求解步骤

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

使用三角恒等式改写

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

使用换元积分法

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

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

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

化简

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

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

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