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 5tan^3(2x)sec^5(2x)

解答

\int 5 tan^{3} (2 x) sec^{5} (2 x) d x

解答

5 (\frac{sec ^{7} ( 2 x )}{14} - \frac{sec ^{5} ( 2 x )}{10}) + C

求解步骤

\int 5 tan^{3} (2 x) sec^{5} (2 x) d x

提出常数:

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

= 5 \cdot \int tan^{3} (2 x) sec^{5} (2 x) d x

使用三角恒等式改写

= 5 \cdot \int (- 1 + sec^{2} (2 x)) tan (2 x) sec^{5} (2 x) d x

使用换元积分法

= 5 \cdot \int \frac{u ^{4} ( u ^{2} - 1 )}{2} d u

乘开

\frac{u ^{4} ( u ^{2} - 1 )}{2} : \frac{u ^{6}}{2} - \frac{u ^{4}}{2}

= 5 \cdot \int \frac{u ^{6}}{2} - \frac{u ^{4}}{2} 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

= 5 (\int \frac{u ^{6}}{2} d u - \int \frac{u ^{4}}{2} d u)

\int \frac{u ^{6}}{2} d u = \frac{u ^{7}}{14}

\int \frac{u ^{4}}{2} d u = \frac{u ^{5}}{10}

= 5 (\frac{u ^{7}}{14} - \frac{u ^{5}}{10})

u = sec (2 x)

代回

= 5 (\frac{sec ^{7} ( 2 x )}{14} - \frac{sec ^{5} ( 2 x )}{10})

解答补常数

= 5 (\frac{sec ^{7} ( 2 x )}{14} - \frac{sec ^{5} ( 2 x )}{10}) + C

作图

流行的例子

s im pl i f y f (g (x))

面积 y=e^x,y=e^{3x},x=0,x=ln(3)

a re a y = e^{x}, y = e^{3 x}, x = 0, x = ln (3)

limit as x approaches 3 of ln(3-x)

x \to 3 lim (ln (3 - x))

derivative e^{11x}

d er i v a t i v e e^{11 x}

integral of 2e^x

\int 2 e^{x} d x