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人気のある微分積分 >

integral of x^2arcsec(x)

解

\int x^{2} arcsec (x) d x

解

\frac{1}{3} x^{3} arcsec (x) - \frac{1}{3} (\frac{1}{2} x x^{2} - 1 + \frac{1}{2} ln x^{2} - 1 + x) + C

解答ステップ

\int x^{2} arcsec (x) d x

u置換積分法を適用する

= \int u sec^{3} (u) tan (u) d u

部分積分を適用する

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

\int \frac{1}{3} sec^{3} (u) d u = \frac{1}{3} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

= \frac{1}{3} u sec^{3} (u) - \frac{1}{3} (\frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

代用を戻す

u = arcsec (x)

= \frac{1}{3} arcsec (x) sec^{3} (arcsec (x)) - \frac{1}{3} (\frac{1}{2} sec (arcsec (x)) tan (arcsec (x)) + \frac{1}{2} ln ∣ tan (arcsec (x)) + sec (arcsec (x)) ∣)

簡素化

\frac{1}{3} arcsec (x) sec^{3} (arcsec (x)) - \frac{1}{3} (\frac{1}{2} sec (arcsec (x)) tan (arcsec (x)) + \frac{1}{2} ln ∣ tan (arcsec (x)) + sec (arcsec (x)) ∣) : \frac{1}{3} x^{3} arcsec (x) - \frac{1}{3} (\frac{1}{2} x x^{2} - 1 + \frac{1}{2} ln x^{2} - 1 + x)

= \frac{1}{3} x^{3} arcsec (x) - \frac{1}{3} (\frac{1}{2} x x^{2} - 1 + \frac{1}{2} ln x^{2} - 1 + x)

定数を解答に追加する

= \frac{1}{3} x^{3} arcsec (x) - \frac{1}{3} (\frac{1}{2} x x^{2} - 1 + \frac{1}{2} ln x^{2} - 1 + x) + C

グラフ

人気の例

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