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人気のある微分積分 >

integral of θarcsec(9θ)

解

\int θ arcsec (9 θ) d θ

解

\frac{1}{162} (81 θ^{2} arcsec (9 θ) - 81 θ^{2} - 1) + C

解答ステップ

\int θ arcsec (9 θ) d θ

u置換積分法を適用する

= \int \frac{u arcsec ( u )}{81} d u

定数を除く:

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

= \frac{1}{81} \cdot \int u arcsec (u) d u

部分積分を適用する

= \frac{1}{81} (\frac{1}{2} u^{2} arcsec (u) - \int \frac{u ^{2}}{2 u ^{2} u ^{2} - 1} d u)

\int \frac{u ^{2}}{2 u ^{2} u ^{2} - 1} d u = \frac{1}{2} u^{2} - 1

= \frac{1}{81} (\frac{1}{2} u^{2} arcsec (u) - \frac{1}{2} u^{2} - 1)

代用を戻す

u = 9 θ

= \frac{1}{81} (\frac{1}{2} (9 θ)^{2} arcsec (9 θ) - \frac{1}{2} (9 θ)^{2} - 1)

簡素化

\frac{1}{81} (\frac{1}{2} (9 θ)^{2} arcsec (9 θ) - \frac{1}{2} (9 θ)^{2} - 1) : \frac{1}{162} (81 θ^{2} arcsec (9 θ) - 81 θ^{2} - 1)

= \frac{1}{162} (81 θ^{2} arcsec (9 θ) - 81 θ^{2} - 1)

定数を解答に追加する

= \frac{1}{162} (81 θ^{2} arcsec (9 θ) - 81 θ^{2} - 1) + C

グラフ

人気の例

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