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人気のある微分積分 >

integral of csc^3(2t-3)

解

\int csc^{3} (2 t - 3) d t

解

\frac{1}{16} (- cot^{2} (\frac{2 t - 3}{2}) + 4 ln tan (\frac{2 t - 3}{2}) + tan^{2} (\frac{2 t - 3}{2})) + C

解答ステップ

\int csc^{3} (2 t - 3) d t

u置換積分法を適用する

= \int csc^{3} (u) \frac{1}{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 csc^{3} (u) d u

u置換積分法を適用する

= \frac{1}{2} \cdot \int \frac{( 1 + v ^{2} ) ^{2}}{4 v ^{3}} d v

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}{4} \cdot \int \frac{( 1 + v ^{2} ) ^{2}}{v ^{3}} d v

拡張

\frac{( 1 + v ^{2} ) ^{2}}{v ^{3}} : \frac{1}{v ^{3}} + \frac{2}{v} + v

= \frac{1}{2} \cdot \frac{1}{4} \cdot \int \frac{1}{v ^{3}} + \frac{2}{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}{2} \cdot \frac{1}{4} (\int \frac{1}{v ^{3}} d v + \int \frac{2}{v} d v + \int v d v)

\int \frac{1}{v ^{3}} d v = - \frac{1}{2 v ^{2}}

\int \frac{2}{v} d v = 2 ln ∣ v ∣

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

= \frac{1}{2} \cdot \frac{1}{4} (- \frac{1}{2 v ^{2}} + 2 ln ∣ v ∣ + \frac{v ^{2}}{2})

代用を戻す

= \frac{1}{2} \cdot \frac{1}{4} (- \frac{1}{2 tan ^{2} ( \frac{2 t - 3}{2} )} + 2 ln tan (\frac{2 t - 3}{2}) + \frac{tan ^{2} ( \frac{2 t - 3}{2} )}{2})

簡素化

\frac{1}{2} \cdot \frac{1}{4} (- \frac{1}{2 tan ^{2} ( \frac{2 t - 3}{2} )} + 2 ln tan (\frac{2 t - 3}{2}) + \frac{tan ^{2} ( \frac{2 t - 3}{2} )}{2}) : \frac{1}{16} (- cot^{2} (\frac{2 t - 3}{2}) + 4 ln tan (\frac{2 t - 3}{2}) + tan^{2} (\frac{2 t - 3}{2}))

= \frac{1}{16} (- cot^{2} (\frac{2 t - 3}{2}) + 4 ln tan (\frac{2 t - 3}{2}) + tan^{2} (\frac{2 t - 3}{2}))

定数を解答に追加する

= \frac{1}{16} (- cot^{2} (\frac{2 t - 3}{2}) + 4 ln tan (\frac{2 t - 3}{2}) + tan^{2} (\frac{2 t - 3}{2})) + C

グラフ

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