zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的微积分 >

integral of 1/(pisqrt(x(1-x)))

解答

\int \frac{1}{π x ( 1 - x )} d x

解答

\frac{1}{π} arcsin (2 x - 1) + C

求解步骤

\int \frac{1}{π x ( 1 - x )} d x

提出常数:

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

= \frac{1}{π} \cdot \int \frac{1}{x ( 1 - x )} d x

乘开

x (1 - x) : x - x^{2}

= \frac{1}{π} \cdot \int \frac{1}{x - x ^{2}} d x

对

x - x^{2}

配方:

- (x - \frac{1}{2})^{2} + \frac{1}{4}

= \frac{1}{π} \cdot \int \frac{1}{- ( x - \frac{1}{2} ) ^{2} + \frac{1}{4}} d x

使用换元积分法

= \frac{1}{π} \cdot \int \frac{2}{- 4 u ^{2} + 1} d u

提出常数:

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

= \frac{1}{π} \cdot 2 \cdot \int \frac{1}{- 4 u ^{2} + 1} d u

使用三角换元法

= \frac{1}{π} \cdot 2 \cdot \int \frac{1}{2} d v

提出常数:

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

= \frac{1}{π} \cdot 2 \cdot \frac{1}{2} \cdot \int 1 d v

常数积分:

\int a d x = a x

= \frac{1}{π} \cdot 2 \cdot \frac{1}{2} \cdot 1 \cdot v

代回

= \frac{1}{π} \cdot 2 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (2 (x - \frac{1}{2}))

化简

\frac{1}{π} \cdot 2 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (2 (x - \frac{1}{2})) : \frac{1}{π} arcsin (2 x - 1)

= \frac{1}{π} arcsin (2 x - 1)

解答补常数

= \frac{1}{π} arcsin (2 x - 1) + C

作图

流行的例子

derivative of cos(2-7x)

\frac{d}{d x} (cos (2 - 7 x))

derivative f(x)=(2x-8)^4(x^2+x+1)^5

d er i v a t i v e f (x) = (2 x - 8)^{4} (x^{2} + x + 1)^{5}

integral from 0 to infinity of te^{-st}

\int_{0}^{\infty} t e^{- s t} d t

integral from 0 to 3 of e^xsin(x)

\int_{0}^{3} e^{x} sin (x) d x

limit as h approaches 0 of 1/(h+1)

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