integral from a to b of sqrt(3-x^2)

解

\int_{a}^{b} 3 - x^{2} d x

\frac{3}{2} (arcsin (\frac{b}{3}) - arcsin (\frac{a}{3}) + \frac{1}{2} (sin (2 arcsin (\frac{b}{3})) - sin (2 arcsin (\frac{a}{3}))))

解答ステップ

\int_{a}^{b} 3 - x^{2} d x

三角関数による置換を適用する

= \int_{a r c s i n (\frac{a}{3})}^{a r c s i n (\frac{b}{3})} 3 cos^{2} (u) d u

定数を除く:

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

= 3 \cdot \int_{a r c s i n (\frac{a}{3})}^{a r c s i n (\frac{b}{3})} cos^{2} (u) d u

三角関数の公式を使用して書き換える

= 3 \cdot \int_{a r c s i n (\frac{a}{3})}^{a r c s i n (\frac{b}{3})} \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

= 3 \cdot \frac{1}{2} \cdot \int_{a r c s i n (\frac{a}{3})}^{a r c s i n (\frac{b}{3})} 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

= 3 \cdot \frac{1}{2} (\int_{a r c s i n (\frac{a}{3})}^{a r c s i n (\frac{b}{3})} 1 d u + \int_{a r c s i n (\frac{a}{3})}^{a r c s i n (\frac{b}{3})} cos (2 u) d u)

\int_{a r c s i n (\frac{a}{3})}^{a r c s i n (\frac{b}{3})} 1 d u = arcsin (\frac{b}{3}) - arcsin (\frac{a}{3})

\int_{a r c s i n (\frac{a}{3})}^{a r c s i n (\frac{b}{3})} cos (2 u) d u = \frac{1}{2} (sin (2 arcsin (\frac{b}{3})) - sin (2 arcsin (\frac{a}{3})))

= 3 \cdot \frac{1}{2} (arcsin (\frac{b}{3}) - arcsin (\frac{a}{3}) + \frac{1}{2} (sin (2 arcsin (\frac{b}{3})) - sin (2 arcsin (\frac{a}{3}))))

簡素化

3 \cdot \frac{1}{2} (arcsin (\frac{b}{3}) - arcsin (\frac{a}{3}) + \frac{1}{2} (sin (2 arcsin (\frac{b}{3})) - sin (2 arcsin (\frac{a}{3})))) : \frac{3}{2} (arcsin (\frac{b}{3}) - arcsin (\frac{a}{3}) + \frac{1}{2} (sin (2 arcsin (\frac{b}{3})) - sin (2 arcsin (\frac{a}{3}))))

= \frac{3}{2} (arcsin (\frac{b}{3}) - arcsin (\frac{a}{3}) + \frac{1}{2} (sin (2 arcsin (\frac{b}{3})) - sin (2 arcsin (\frac{a}{3}))))