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人気のある微分積分 >

integral of tcos^5(t^2)

解

\int t cos^{5} (t^{2}) d t

解

\frac{1}{2} sin (t^{2}) - \frac{1}{3} sin^{3} (t^{2}) + \frac{1}{10} sin^{5} (t^{2}) + C

解答ステップ

\int t cos^{5} (t^{2}) d t

u置換積分法を適用する

= \int \frac{cos ^{5} ( u )}{2} 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 \int cos^{5} (u) d u

三角関数の公式を使用して書き換える

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

u置換積分法を適用する

= \frac{1}{2} \cdot \int (1 - v^{2})^{2} d v

拡張

(1 - v^{2})^{2} : 1 - 2 v^{2} + v^{4}

= \frac{1}{2} \cdot \int 1 - 2 v^{2} + v^{4} d v

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 1 d v - \int 2 v^{2} d v + \int v^{4} d v)

\int 1 d v = v

\int 2 v^{2} d v = \frac{2 v ^{3}}{3}

\int v^{4} d v = \frac{v ^{5}}{5}

= \frac{1}{2} (v - \frac{2 v ^{3}}{3} + \frac{v ^{5}}{5})

代用を戻す

= \frac{1}{2} (sin (t^{2}) - \frac{2 sin ^{3} ( t ^{2} )}{3} + \frac{sin ^{5} ( t ^{2} )}{5})

簡素化

\frac{1}{2} (sin (t^{2}) - \frac{2 sin ^{3} ( t ^{2} )}{3} + \frac{sin ^{5} ( t ^{2} )}{5}) : \frac{1}{2} sin (t^{2}) - \frac{1}{3} sin^{3} (t^{2}) + \frac{1}{10} sin^{5} (t^{2})

= \frac{1}{2} sin (t^{2}) - \frac{1}{3} sin^{3} (t^{2}) + \frac{1}{10} sin^{5} (t^{2})

定数を解答に追加する

= \frac{1}{2} sin (t^{2}) - \frac{1}{3} sin^{3} (t^{2}) + \frac{1}{10} sin^{5} (t^{2}) + C

グラフ

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