integral of sqrt(1-z^2)

解答

\int 1 - z^{2}

\frac{1}{2} (arcsin (z) + \frac{1}{2} sin (2 arcsin (z))) + C

求解步骤

\int 1 - z^{2} d z

使用三角换元法

= \int cos^{2} (u) d u

使用三角恒等式改写

= \int \frac{1 + cos ( 2 u )}{2} d u

提出常数:

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

= \frac{1}{2} \cdot \int 1 + cos (2 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

= \frac{1}{2} (\int 1 d u + \int cos (2 u) d u)

\int 1 d u = u

\int cos (2 u) d u = \frac{1}{2} sin (2 u)

= \frac{1}{2} (u + \frac{1}{2} sin (2 u))

u = arcsin (z)

代回

= \frac{1}{2} (arcsin (z) + \frac{1}{2} sin (2 arcsin (z)))

解答补常数

= \frac{1}{2} (arcsin (z) + \frac{1}{2} sin (2 arcsin (z))) + C

h \to 0 lim (\frac{1}{h})

x \to 0 lim (\frac{x}{sin ( 5 x )})

\int_{0}^{4} \frac{1}{x ^{2} - x - 2} d x

\int_{0}^{π} x^{2} cos (n x) d x

\frac{d}{d x} (x^{3} + 3 x^{2} - 24 x - 80)