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人気のある微分積分 >

integral of (tan^3(9/z))/(z^2)

解

\int \frac{tan ^{3} ( \frac{9}{z} )}{z ^{2}}

解

- \frac{1}{9} (- ln sec (\frac{9}{z}) + \frac{sec ^{2} ( \frac{9}{z} )}{2}) + C

解答ステップ

\int \frac{tan ^{3} ( \frac{9}{z} )}{z ^{2}} d z

u置換積分法を適用する

= \int - \frac{tan ^{3} ( u )}{9} d u

定数を除く:

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

= - \frac{1}{9} \cdot \int tan^{3} (u) d u

三角関数の公式を使用して書き換える

= - \frac{1}{9} \cdot \int (- 1 + sec^{2} (u)) tan (u) d u

u置換積分法を適用する

= - \frac{1}{9} \cdot \int \frac{- 1 + v ^{2}}{v} d v

拡張

\frac{- 1 + v ^{2}}{v} : - \frac{1}{v} + v

= - \frac{1}{9} \cdot \int - \frac{1}{v} + v 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}{9} (- \int \frac{1}{v} d v + \int v d v)

\int \frac{1}{v} d v = ln ∣ v ∣

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

= - \frac{1}{9} (- ln ∣ v ∣ + \frac{v ^{2}}{2})

代用を戻す

= - \frac{1}{9} (- ln sec (\frac{9}{z}) + \frac{sec ^{2} ( \frac{9}{z} )}{2})

定数を解答に追加する

= - \frac{1}{9} (- ln sec (\frac{9}{z}) + \frac{sec ^{2} ( \frac{9}{z} )}{2}) + C

グラフ

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