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人気のある微分積分 >

integral of 8sin^2(t+pi/(12))

解

\int 8 sin^{2} (t + \frac{π}{12}) d t

解

4 (t + \frac{π}{12} - \frac{1}{2} sin (2 (t + \frac{π}{12}))) + C

解答ステップ

\int 8 sin^{2} (t + \frac{π}{12}) d t

定数を除く:

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

= 8 \cdot \int sin^{2} (t + \frac{π}{12}) d t

u置換積分法を適用する

= 8 \cdot \int sin^{2} (u) d u

三角関数の公式を使用して書き換える

= 8 \cdot \int \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

= 8 \cdot \frac{1}{2} \cdot \int 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

= 8 \cdot \frac{1}{2} (\int 1 d u - \int cos (2 u) d u)

\int 1 d u = u

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

= 8 \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

代用を戻す

u = t + \frac{π}{12}

= 8 \cdot \frac{1}{2} (t + \frac{π}{12} - \frac{1}{2} sin (2 (t + \frac{π}{12})))

簡素化

8 \cdot \frac{1}{2} (t + \frac{π}{12} - \frac{1}{2} sin (2 (t + \frac{π}{12}))) : 4 (t + \frac{π}{12} - \frac{1}{2} sin (2 (t + \frac{π}{12})))

= 4 (t + \frac{π}{12} - \frac{1}{2} sin (2 (t + \frac{π}{12})))

定数を解答に追加する

= 4 (t + \frac{π}{12} - \frac{1}{2} sin (2 (t + \frac{π}{12}))) + C

グラフ

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