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 e^{cos(2x)}sin(x)cos(x)

解答

\int e^{c o s (2 x)} sin (x) cos (x) d x

解答

- \frac{1}{4} e^{c o s (2 x)} + C

求解步骤

\int e^{c o s (2 x)} sin (x) cos (x) d x

使用三角恒等式改写

= \int \frac{1}{2} e^{c o s (2 x)} sin (2 x) d x

提出常数:

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

= \frac{1}{2} \cdot \int e^{c o s (2 x)} sin (2 x) d x

使用换元积分法

= \frac{1}{2} \cdot \int e^{c o s (u)} sin (u) \frac{1}{2} d u

提出常数:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int e^{c o s (u)} sin (u) d u

使用换元积分法

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

提出常数:

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

= \frac{1}{2} \cdot \frac{1}{2} (- \int e^{v} d v)

使用积分常见定则:

\int e^{v} d v = e^{v}

= \frac{1}{2} \cdot \frac{1}{2} (- e^{v})

代回

= \frac{1}{2} \cdot \frac{1}{2} (- e^{c o s (2 x)})

化简

\frac{1}{2} \cdot \frac{1}{2} (- e^{c o s (2 x)}) : - \frac{1}{4} e^{c o s (2 x)}

= - \frac{1}{4} e^{c o s (2 x)}

解答补常数

= - \frac{1}{4} e^{c o s (2 x)} + C

作图

流行的例子

(t^2+x^2)dt+(2tx-x)dx=0

(t^{2} + x^{2}) d t + (2 t x - x) d x = 0

limit as x approaches 0-of-7x^2+5x

x \to 0 - lim (- 7 x^{2} + 5 x)

derivative y=(6x-1)/(8x+3)

d er i v a t i v e y = \frac{6 x - 1}{8 x + 3}

泰勒 tan^{-1}(x)

t a y l or tan^{- 1} (x)

derivative of (2-6x^{0.5})

\frac{d}{d x} ((2 - 6 x)^{0.5})