integral from 0 to x of sqrt(36-t^2)

解

\int_{0}^{x} 36 - t^{2} d t

18 (arcsin (\frac{x}{6}) + \frac{1}{2} sin (2 arcsin (\frac{x}{6})))

解答ステップ

\int_{0}^{x} 36 - t^{2} d t

三角関数による置換を適用する

= \int_{0}^{a r c s i n (\frac{x}{6})} 36 cos^{2} (u) d u

定数を除く:

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

= 36 \cdot \int_{0}^{a r c s i n (\frac{x}{6})} cos^{2} (u) d u

三角関数の公式を使用して書き換える

= 36 \cdot \int_{0}^{a r c s i n (\frac{x}{6})} \frac{1 + cos ( 2 u )}{2} d u

定数を除く:

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

= 36 \cdot \frac{1}{2} \cdot \int_{0}^{a r c s i n (\frac{x}{6})} 1 + cos (2 u) 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

= 36 \cdot \frac{1}{2} (\int_{0}^{a r c s i n (\frac{x}{6})} 1 d u + \int_{0}^{a r c s i n (\frac{x}{6})} cos (2 u) d u)

\int_{0}^{a r c s i n (\frac{x}{6})} 1 d u = arcsin (\frac{x}{6})

\int_{0}^{a r c s i n (\frac{x}{6})} cos (2 u) d u = \frac{1}{2} sin (2 arcsin (\frac{x}{6}))

= 36 \cdot \frac{1}{2} (arcsin (\frac{x}{6}) + \frac{1}{2} sin (2 arcsin (\frac{x}{6})))

簡素化

36 \cdot \frac{1}{2} (arcsin (\frac{x}{6}) + \frac{1}{2} sin (2 arcsin (\frac{x}{6}))) : 18 (arcsin (\frac{x}{6}) + \frac{1}{2} sin (2 arcsin (\frac{x}{6})))

= 18 (arcsin (\frac{x}{6}) + \frac{1}{2} sin (2 arcsin (\frac{x}{6})))