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/(sin(x)-2csc(x))

解答

\int \frac{1}{sin ( x ) - 2 csc ( x )} d x

解答

arctan (2 tan (\frac{x}{2}) + 1) - arctan (2 tan (\frac{x}{2}) - 1) + C

求解步骤

\int \frac{1}{sin ( x ) - 2 csc ( x )} d x

使用换元积分法

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

提出常数:

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

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

将

\frac{u}{- u ^{4} - 1}

用部份分式展开:

\frac{1}{2 2 ( u ^{2} + 2 u + 1 )} - \frac{1}{2 2 ( u ^{2} - 2 u + 1 )}

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

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

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

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

= 2 (\frac{1}{2} arctan (2 u + 1) - \frac{1}{2} arctan (2 u - 1))

u = tan (\frac{x}{2})

代回

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

化简

2 (\frac{1}{2} arctan (2 tan (\frac{x}{2}) + 1) - \frac{1}{2} arctan (2 tan (\frac{x}{2}) - 1)) : arctan (2 tan (\frac{x}{2}) + 1) - arctan (2 tan (\frac{x}{2}) - 1)

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

解答补常数

= arctan (2 tan (\frac{x}{2}) + 1) - arctan (2 tan (\frac{x}{2}) - 1) + C

作图

流行的例子

tangent y=(5x)/(x-2),(3,15)

t an g e n t y = \frac{5 x}{x - 2}, (3, 15)

斜率 f(x)=(x^2)/(x+4)

s l o p e f (x) = \frac{x ^{2}}{x + 4}

integral of (6e^x)

\int (6 e^{x}) d x

sum from n=0 to infinity of 7/(2^n)

n = 0 \sum \infty \frac{7}{2 ^{n}}

derivative of e^xx^6+12e^xx^5+30e^xx^4

\frac{d}{d x} (e^{x} x^{6} + 12 e^{x} x^{5} + 30 e^{x} x^{4})