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人気のある微分積分 >

integral of x^4tan^3(x^5)

解

\int x^{4} tan^{3} (x^{5}) d x

解

\frac{1}{10} (- ln (sec^{2} (x^{5})) + sec^{2} (x^{5})) + C

解答ステップ

\int x^{4} tan^{3} (x^{5}) d x

u置換積分法を適用する

= \int \frac{3 u}{15 ( u ^{\frac{2}{3}} + 1 )} d u

定数を除く:

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

= \frac{1}{15} \cdot \int \frac{3 u}{u ^{\frac{2}{3}} + 1} d u

u置換積分法を適用する

= \frac{1}{15} \cdot \int \frac{3 v ^{3}}{v ^{2} + 1} d v

定数を除く:

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

= \frac{1}{15} \cdot 3 \cdot \int \frac{v ^{3}}{v ^{2} + 1} d v

u置換積分法を適用する

= \frac{1}{15} \cdot 3 \cdot \int \frac{w - 1}{2 w} d w

定数を除く:

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

= \frac{1}{15} \cdot 3 \cdot \frac{1}{2} \cdot \int \frac{w - 1}{w} d w

拡張

\frac{w - 1}{w} : 1 - \frac{1}{w}

= \frac{1}{15} \cdot 3 \cdot \frac{1}{2} \cdot \int 1 - \frac{1}{w} d w

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}{15} \cdot 3 \cdot \frac{1}{2} (\int 1 d w - \int \frac{1}{w} d w)

\int 1 d w = w

\int \frac{1}{w} d w = ln ∣ w ∣

= \frac{1}{15} \cdot 3 \cdot \frac{1}{2} (w - ln ∣ w ∣)

代用を戻す

= \frac{1}{15} \cdot 3 \cdot \frac{1}{2} ((3 tan^{3} (x^{5}))^{2} + 1 - ln (3 tan^{3} (x^{5}))^{2} + 1)

簡素化

\frac{1}{15} \cdot 3 \cdot \frac{1}{2} ((3 tan^{3} (x^{5}))^{2} + 1 - ln (3 tan^{3} (x^{5}))^{2} + 1) : \frac{1}{10} (- ln (sec^{2} (x^{5})) + sec^{2} (x^{5}))

= \frac{1}{10} (- ln (sec^{2} (x^{5})) + sec^{2} (x^{5}))

定数を解答に追加する

= \frac{1}{10} (- ln (sec^{2} (x^{5})) + sec^{2} (x^{5})) + C

グラフ

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