integral from 0 to 1 of sin^3(x)

解

\int_{0}^{1} sin^{3} (x) d x

1 - cos (1) - \frac{1 - cos ^{3} ( 1 )}{3}

十進法表記

0.17894 \dots

解答ステップ

\int_{0}^{1} sin^{3} (x) d x

三角関数の公式を使用して書き換える

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

u置換積分法を適用する

= \int_{1}^{c o s (1)} - 1 + u^{2} d u

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

= - \int_{c o s (1)}^{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

= - (- \int_{c o s (1)}^{1} 1 d u + \int_{c o s (1)}^{1} u^{2} d u)

\int_{c o s (1)}^{1} 1 d u = 1 - cos (1)

\int_{c o s (1)}^{1} u^{2} d u = \frac{1 - cos ^{3} ( 1 )}{3}

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

簡素化

= 1 - cos (1) - \frac{1 - cos ^{3} ( 1 )}{3}