integral of (sin^2(x))/(sqrt(x))

解

\int \frac{sin ^{2} ( x )}{x} d x

\frac{1}{2} (2 x - π C (2 \frac{2}{π} x)) + C

解答ステップ

\int \frac{sin ^{2} ( x )}{x} d x

三角関数の公式を使用して書き換える

= \int \frac{1 - cos ( 2 x )}{2 x} d x

定数を除く:

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

= \frac{1}{2} \cdot \int \frac{1 - cos ( 2 x )}{x} d x

u置換積分法を適用する

= \frac{1}{2} \cdot \int \frac{1 - cos ( u )}{2 u} d u

定数を除く:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int \frac{1 - cos ( u )}{u} d u

拡張

\frac{1 - cos ( u )}{u} : \frac{1}{u} - \frac{cos ( u )}{u}

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int \frac{1}{u} - \frac{cos ( u )}{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

= \frac{1}{2} \cdot \frac{1}{2} (\int \frac{1}{u} d u - \int \frac{cos ( u )}{u} d u)

\int \frac{1}{u} d u = 2 u

\int \frac{cos ( u )}{u} d u = 2 π C (\frac{2}{π} u)

= \frac{1}{2} \cdot \frac{1}{2} (2 u - 2 π C (\frac{2}{π} u))

代用を戻す

u = 2 x

= \frac{1}{2} \cdot \frac{1}{2} (2 2 x - 2 π C (\frac{2}{π} 2 x))

簡素化

\frac{1}{2} \cdot \frac{1}{2} (2 2 x - 2 π C (\frac{2}{π} 2 x)) : \frac{1}{2} (2 x - π C (2 \frac{2}{π} x))

= \frac{1}{2} (2 x - π C (2 \frac{2}{π} x))

定数を解答に追加する

= \frac{1}{2} (2 x - π C (2 \frac{2}{π} x)) + C