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人気のある微分積分 >

integral from 0 to pi/2 of (sin^5(θ))/4

解

\int_{0}^{\frac{π}{2}} \frac{sin ^{5} ( θ )}{4} d θ

解

\frac{2}{15}

+1

十進法表記

0.13333 \dots

解答ステップ

\int_{0}^{\frac{π}{2}} \frac{sin ^{5} ( θ )}{4} d θ

定数を除く:

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

= \frac{1}{4} \cdot \int_{0}^{\frac{π}{2}} sin^{5} (θ) d θ

三角関数の公式を使用して書き換える

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

u置換積分法を適用する

= \frac{1}{4} \cdot \int_{1}^{0} - 1 + 2 u^{2} - u^{4} d u

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

= \frac{1}{4} (- \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

= \frac{1}{4} (- (- \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}

= \frac{1}{4} (- (- 1 + \frac{2}{3} - \frac{1}{5}))

簡素化

\frac{1}{4} (- (- 1 + \frac{2}{3} - \frac{1}{5})) : \frac{2}{15}

= \frac{2}{15}

グラフ

人気の例

integral from 0 to 3 of (x^3)/3

\int_{0}^{3} \frac{x ^{3}}{3} d x

integral from 2 to 5 of (2+3x)

\int_{2}^{5} (2 + 3 x) d x

integral from 1 to 3 of (2x^2-x)

\int_{1}^{3} (2 x^{2} - x) d x

integral from-1 to 1 of (2x-4y)

\int_{- 1}^{1} (2 x - 4 y) d y

integral from 0 to 1 of 1/((1+3y)^2)

\int_{0}^{1} \frac{1}{( 1 + 3 y ) ^{2}} d y