integral from 0 to pi/2 of 3sin^3(x)

解

\int_{0}^{\frac{π}{2}} 3 sin^{3} (x) d x

2

解答ステップ

\int_{0}^{\frac{π}{2}} 3 sin^{3} (x) d x

定数を除く:

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

= 3 \cdot \int_{0}^{\frac{π}{2}} sin^{3} (x) d x

三角関数の公式を使用して書き換える

= 3 \cdot \int_{0}^{\frac{π}{2}} (1 - cos^{2} (x)) sin (x) d x

u置換積分法を適用する

= 3 \cdot \int_{1}^{0} - 1 + u^{2} d u

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

= 3 (- \int_{0}^{1} - 1 + u^{2} 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

= 3 (- (- \int_{0}^{1} 1 d u + \int_{0}^{1} u^{2} d u))

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

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

= 3 (- (- 1 + \frac{1}{3}))

簡素化

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

= 2