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受欢迎的微积分 >

integral of sqrt(y^2-1)

解答

\int y^{2} - 1 d y

解答

\frac{1}{2} y y^{2} - 1 - \frac{1}{2} ln y + y^{2} - 1 + C

求解步骤

\int y^{2} - 1 d y

使用三角换元法

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

使用三角恒等式改写

= \int (- 1 + sec^{2} (u)) sec (u) d u

乘开

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

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

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

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

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

u = arcsec (y)

代回

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

化简

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

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

解答补常数

= \frac{1}{2} y y^{2} - 1 - \frac{1}{2} ln y + y^{2} - 1 + C

作图

流行的例子

tangent f(x)=x^3-2x+2,(4,58)

t an g e n t f (x) = x^{3} - 2 x + 2, (4, 58)

laplacetransform e^{-t}(cos(4t)-2sin(4t))

l a pl a ce t r an s f or m e^{- t} (cos (4 t) - 2 sin (4 t))

\frac{d}{d t} (t^{2} e^{t})

泰勒 3/((x^2-x-2))

t a y l or \frac{3}{( x ^{2} - x - 2 )}

derivative of (2+x^{1/3}(1-x)^{2/3})

\frac{d}{d x} ((2 + x)^{\frac{1}{3}} (1 - x)^{\frac{2}{3}})