integral from 1 to x of sqrt(t^2-1)

解答

\int_{1}^{x} t^{2} - 1 d t

- ln x + x^{2} - 1 + \frac{x x ^{2} - 1 + ln x + x ^{2} - 1}{2}

求解步骤

\int_{1}^{x} t^{2} - 1 d t

使用三角换元法

= \int_{0}^{arcsec (x)} tan^{2} (u) sec (u) d u

使用三角恒等式改写

= \int_{0}^{arcsec (x)} (- 1 + sec^{2} (u)) sec (u) d u

乘开

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

= \int_{0}^{arcsec (x)} - sec (u) + sec^{3} (u) d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{0}^{arcsec (x)} sec (u) d u + \int_{0}^{arcsec (x)} sec^{3} (u) d u

\int_{0}^{arcsec (x)} sec (u) d u = ln x + x^{2} - 1

\int_{0}^{arcsec (x)} sec^{3} (u) d u = \frac{x x ^{2} - 1 + ln x + x ^{2} - 1}{2}

= - ln x + x^{2} - 1 + \frac{x x ^{2} - 1 + ln x + x ^{2} - 1}{2}

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