integral of (sin(2e^{-4x}))/(e^{4x)}

解答

\int \frac{sin ( 2 e ^{- 4 x} )}{e ^{4 x}} d x

\frac{1}{8} cos (2 e^{- 4 x}) + C

求解步骤

\int \frac{sin ( 2 e ^{- 4 x} )}{e ^{4 x}} d x

使用换元积分法

= \int - \frac{sin ( u )}{8} d u

提出常数:

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

= - \frac{1}{8} \cdot \int sin (u) d u

使用积分常见定则:

\int sin (u) d u = - cos (u)

= - \frac{1}{8} (- cos (u))

u = 2 e^{- 4 x}

代回

= - \frac{1}{8} (- cos (2 e^{- 4 x}))

化简

= \frac{1}{8} cos (2 e^{- 4 x})

解答补常数

= \frac{1}{8} cos (2 e^{- 4 x}) + C

a re a f (x) = x^{3} - 3 x^{2} + 3 x, g (x) = x

d er i v a t i v e f (3) = 2 x^{2} + 4 x

\int \frac{1}{x ^{2} 25 x ^{2} + 64} d x

\frac{\partial}{\partial y} (- 3 x^{2} y)

a re a f (x) = x^{2} + 4 x, g (x) = x + 10