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 tan^3(7x+8)

解答

\int tan^{3} (7 x + 8) d x

解答

\frac{1}{7} (- ln ∣ sec (7 x + 8) ∣ + \frac{sec ^{2} ( 7 x + 8 )}{2}) + C

求解步骤

\int tan^{3} (7 x + 8) d x

使用换元积分法

= \int tan^{3} (u) \frac{1}{7} d u

提出常数:

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

= \frac{1}{7} \cdot \int tan^{3} (u) d u

使用三角恒等式改写

= \frac{1}{7} \cdot \int (- 1 + sec^{2} (u)) tan (u) d u

使用换元积分法

= \frac{1}{7} \cdot \int \frac{- 1 + v ^{2}}{v} d v

乘开

\frac{- 1 + v ^{2}}{v} : - \frac{1}{v} + v

= \frac{1}{7} \cdot \int - \frac{1}{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}{7} (- \int \frac{1}{v} d v + \int v d v)

\int \frac{1}{v} d v = ln ∣ v ∣

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

= \frac{1}{7} (- ln ∣ v ∣ + \frac{v ^{2}}{2})

代回

= \frac{1}{7} (- ln ∣ sec (7 x + 8) ∣ + \frac{sec ^{2} ( 7 x + 8 )}{2})

解答补常数

= \frac{1}{7} (- ln ∣ sec (7 x + 8) ∣ + \frac{sec ^{2} ( 7 x + 8 )}{2}) + C

作图

流行的例子

integral of (2+x+2x^2+e^x)

\int (2 + x + 2 x^{2} + e^{x}) d x

integral of x*cos(x)*sin(x)

\int x \cdot cos (x) \cdot sin (x) d x

integral of ((4+x^2)^{1/2})/(x^6)

\int \frac{( 4 + x ^{2} ) ^{\frac{1}{2}}}{x ^{6}} d x

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

\int \frac{e ^{x}}{( e ^{x} + 2 ) ^{2}} d x

integral of (1+2/(x^2)+3/(x^3))

\int (1 + \frac{2}{x ^{2}} + \frac{3}{x ^{3}}) d x