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受欢迎的微积分 >

integral of 1/(sqrt(1-cos(x)))

解答

\int \frac{1}{1 - cos ( x )} d x

解答

- 2 (\frac{1}{2} ln sec (\frac{x}{2}) + 1 - \frac{1}{2} ln sec (\frac{x}{2}) - 1) + C

求解步骤

\int \frac{1}{1 - cos ( x )} d x

使用换元积分法

= \int \frac{2}{u 1 + u ^{2}} d u

提出常数:

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

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

使用换元积分法

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

因式分解

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

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

提出常数:

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

= 2 (- \int \frac{1}{- v ^{2} + 1} d v)

使用积分常见定则:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 2 (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

代回

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

化简

2 - \frac{ln 1 + tan ^{2} ( \frac{x}{2} ) + 1}{2} - \frac{ln 1 + tan ^{2} ( \frac{x}{2} ) - 1}{2} : - 2 (\frac{1}{2} ln sec (\frac{x}{2}) + 1 - \frac{1}{2} ln sec (\frac{x}{2}) - 1)

= - 2 (\frac{1}{2} ln sec (\frac{x}{2}) + 1 - \frac{1}{2} ln sec (\frac{x}{2}) - 1)

解答补常数

= - 2 (\frac{1}{2} ln sec (\frac{x}{2}) + 1 - \frac{1}{2} ln sec (\frac{x}{2}) - 1) + C

作图

流行的例子

面积 x=-2,x=2,y=2x^2+12,y=0

a re a x = - 2, x = 2, y = 2 x^{2} + 12, y = 0

integral of (\sqrt[3]{x^5})

\int (3 x^{5}) d x

integral of (x+1)/(x^2+1)

\int \frac{x + 1}{x ^{2} + 1} d x

derivative of 8cosh(x/4 ln(arcsin(x)))

\frac{d}{d x} (8 cosh (\frac{x}{4}) ln (arcsin (x)))

(d^2)/(dx^2)(2x^{-7})

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