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 csc^3(5x)

解答

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

解答

\frac{1}{40} (- cot^{2} (\frac{5 x}{2}) + 4 ln tan (\frac{5 x}{2}) + tan^{2} (\frac{5 x}{2})) + C

求解步骤

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

使用换元积分法

= \int csc^{3} (u) \frac{1}{5} d u

提出常数:

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

= \frac{1}{5} \cdot \int csc^{3} (u) d u

使用换元积分法

= \frac{1}{5} \cdot \int \frac{( 1 + v ^{2} ) ^{2}}{4 v ^{3}} d v

提出常数:

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

= \frac{1}{5} \cdot \frac{1}{4} \cdot \int \frac{( 1 + v ^{2} ) ^{2}}{v ^{3}} d v

乘开

\frac{( 1 + v ^{2} ) ^{2}}{v ^{3}} : \frac{1}{v ^{3}} + \frac{2}{v} + v

= \frac{1}{5} \cdot \frac{1}{4} \cdot \int \frac{1}{v ^{3}} + \frac{2}{v} + v d v

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}{5} \cdot \frac{1}{4} (\int \frac{1}{v ^{3}} d v + \int \frac{2}{v} d v + \int v d v)

\int \frac{1}{v ^{3}} d v = - \frac{1}{2 v ^{2}}

\int \frac{2}{v} d v = 2 ln ∣ v ∣

\int v d v = \frac{v ^{2}}{2}

= \frac{1}{5} \cdot \frac{1}{4} (- \frac{1}{2 v ^{2}} + 2 ln ∣ v ∣ + \frac{v ^{2}}{2})

代回

= \frac{1}{5} \cdot \frac{1}{4} (- \frac{1}{2 tan ^{2} ( \frac{5 x}{2} )} + 2 ln tan (\frac{5 x}{2}) + \frac{tan ^{2} ( \frac{5 x}{2} )}{2})

化简

\frac{1}{5} \cdot \frac{1}{4} (- \frac{1}{2 tan ^{2} ( \frac{5 x}{2} )} + 2 ln tan (\frac{5 x}{2}) + \frac{tan ^{2} ( \frac{5 x}{2} )}{2}) : \frac{1}{40} (- cot^{2} (\frac{5 x}{2}) + 4 ln tan (\frac{5 x}{2}) + tan^{2} (\frac{5 x}{2}))

= \frac{1}{40} (- cot^{2} (\frac{5 x}{2}) + 4 ln tan (\frac{5 x}{2}) + tan^{2} (\frac{5 x}{2}))

解答补常数

= \frac{1}{40} (- cot^{2} (\frac{5 x}{2}) + 4 ln tan (\frac{5 x}{2}) + tan^{2} (\frac{5 x}{2})) + C

作图

流行的例子

integral of 1/((x^2-2x+5)^2)

\int \frac{1}{( x ^{2} - 2 x + 5 ) ^{2}} d x

f(x)=-x^4+18x^2-11

f (x) = - x^{4} + 18 x^{2} - 11

derivative of (x^2+7\sqrt[3]{x^2+2})

\frac{d}{d x} ((x^{2} + 7) 3 x^{2} + 2)

integral of-4sin(4t)+1

\int - 4 sin (4 t) + 1 d t

limit as x approaches+4 of 9x+|x-4|

x \to + 4 lim (9 x + ∣ x - 4 ∣)