ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある微分積分 >

integral from 0 to pi/2 of 8cos^5(x)

解

\int_{0}^{\frac{π}{2}} 8 cos^{5} (x) d x

解

\frac{64}{15}

+1

十進法表記

4.26666 \dots

解答ステップ

\int_{0}^{\frac{π}{2}} 8 cos^{5} (x) d x

定数を除く:

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

= 8 \cdot \int_{0}^{\frac{π}{2}} cos^{5} (x) d x

三角関数の公式を使用して書き換える

= 8 \cdot \int_{0}^{\frac{π}{2}} (1 - sin^{2} (x))^{2} cos (x) d x

u置換積分法を適用する

= 8 \cdot \int_{0}^{1} (1 - u^{2})^{2} d u

拡張

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

= 8 \cdot \int_{0}^{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

= 8 (\int_{0}^{1} 1 d u - \int_{0}^{1} 2 u^{2} d u + \int_{0}^{1} u^{4} d u)

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

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

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

= 8 (1 - \frac{2}{3} + \frac{1}{5})

簡素化

8 (1 - \frac{2}{3} + \frac{1}{5}) : \frac{64}{15}

= \frac{64}{15}

グラフ

人気の例

\frac{d y}{d x} = 3 x - y

laplacetransform sin(t+pi/4)

l a pl a ce t r an s f or m sin (t + \frac{π}{4})

integral from 0 to 1 of (1-x^2)^2

\int_{0}^{1} (1 - x^{2})^{2} d x

(2 - x)^{^{'}}

derivative of-7cos(x)

\frac{d}{d x} (- 7 cos (x))