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

解

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

\frac{1}{2} (x - \frac{π}{3} - \frac{1}{2} (sin (2 x) - \frac{3}{2}))

解答ステップ

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

三角関数の公式を使用して書き換える

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

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{π}{3}}^{x} 1 - cos (2 t) d t

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} (\int_{\frac{π}{3}}^{x} 1 d t - \int_{\frac{π}{3}}^{x} cos (2 t) d t)

\int_{\frac{π}{3}}^{x} 1 d t = x - \frac{π}{3}

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

= \frac{1}{2} (x - \frac{π}{3} - \frac{1}{2} (sin (2 x) - \frac{3}{2}))