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人気のある微分積分 >

integral of sqrt(u^2-a^2)

解

\int u^{2} - a^{2} d u

解

\frac{1}{2} u u^{2} - a^{2} - \frac{a ^{2}}{2} ln (\frac{u + u ^{2} - a ^{2}}{∣ a ∣}) + C

解答ステップ

\int u^{2} - a^{2} d u

三角関数による置換を適用する

= \int a^{2} tan^{2} (v) sec (v) d v

定数を除く:

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

= a^{2} \cdot \int tan^{2} (v) sec (v) d v

三角関数の公式を使用して書き換える

= a^{2} \cdot \int (- 1 + sec^{2} (v)) sec (v) d v

拡張

(- 1 + sec^{2} (v)) sec (v) : - sec (v) + sec^{3} (v)

= a^{2} \cdot \int - sec (v) + sec^{3} (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

= a^{2} (- \int sec (v) d v + \int sec^{3} (v) d v)

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

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

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

代用を戻す

v = arcsec (\frac{1}{a} u)

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

簡素化

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

= \frac{1}{2} u u^{2} - a^{2} - \frac{a ^{2}}{2} ln (\frac{u + u ^{2} - a ^{2}}{∣ a ∣})

定数を解答に追加する

= \frac{1}{2} u u^{2} - a^{2} - \frac{a ^{2}}{2} ln (\frac{u + u ^{2} - a ^{2}}{∣ a ∣}) + C

グラフ

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