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^6(6x)

解答

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

解答

\frac{1}{6} (\frac{tan ^{5} ( 6 x )}{5} - \frac{tan ^{3} ( 6 x )}{3} + tan (6 x) - arctan (tan (6 x))) + C

求解步骤

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

使用换元积分法

= \int tan^{6} (u) \frac{1}{6} d u

提出常数:

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

= \frac{1}{6} \cdot \int tan^{6} (u) d u

使用换元积分法

= \frac{1}{6} \cdot \int \frac{v ^{6}}{1 + v ^{2}} d v

长除

\frac{v ^{6}}{1 + v ^{2}} : v^{4} - v^{2} + 1 - \frac{1}{v ^{2} + 1}

= \frac{1}{6} \cdot \int v^{4} - v^{2} + 1 - \frac{1}{v ^{2} + 1} 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}{6} (\int v^{4} d v - \int v^{2} d v + \int 1 d v - \int \frac{1}{v ^{2} + 1} d v)

\int v^{4} d v = \frac{v ^{5}}{5}

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

\int 1 d v = v

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

= \frac{1}{6} (\frac{v ^{5}}{5} - \frac{v ^{3}}{3} + v - arctan (v))

代回

= \frac{1}{6} (\frac{tan ^{5} ( 6 x )}{5} - \frac{tan ^{3} ( 6 x )}{3} + tan (6 x) - arctan (tan (6 x)))

解答补常数

= \frac{1}{6} (\frac{tan ^{5} ( 6 x )}{5} - \frac{tan ^{3} ( 6 x )}{3} + tan (6 x) - arctan (tan (6 x))) + C

作图

流行的例子

integral of sec(x+pi/2)tan(x+pi/2)

\int sec (x + \frac{π}{2}) tan (x + \frac{π}{2}) d x

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

\int \frac{1}{y ^{2} - 2 y + 5} d y

integral of 1/(2x^2-18)

\int \frac{1}{2 x ^{2} - 18} d x

integral of 2x^3(x^2+1)^{65}

\int 2 x^{3} (x^{2} + 1)^{65} d x

integral of 1/(17x)

\int \frac{1}{17 x} d x