What is the integral of e^{-2x}cos(e^{-x}) ?

The integral of e^{-2x}cos(e^{-x}) is -e^{-x}sin(e^{-x})-cos(e^{-x})+C

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integral of e^{-2x}cos(e^{-x})

\int e^{- 2 x} cos (e^{- x}) d x

- e^{- x} sin (e^{- x}) - cos (e^{- x}) + C

Solution steps

\int e^{- 2 x} cos (e^{- x}) d x

Apply u-substitution

: \int - \frac{cos ( u )}{2} d u

= \int - \frac{cos ( u )}{2} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int cos (u) d u

Apply u-substitution

: \int cos (v) \cdot 2 v d v

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

Take the constant out:

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

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

Apply Integration By Parts

: v sin (v) - \int sin (v) d v

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

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

= - \frac{1}{2} \cdot 2 (v sin (v) - (- cos (v)))

Substitute back

= - \frac{1}{2} \cdot 2 (e^{- 2 x} sin (e^{- 2 x}) - (- cos (e^{- 2 x})))

Simplify

- \frac{1}{2} \cdot 2 (e^{- 2 x} sin (e^{- 2 x}) - (- cos (e^{- 2 x}))) : - e^{- x} sin (e^{- x}) - cos (e^{- x})

= - e^{- x} sin (e^{- x}) - cos (e^{- x})

Add a constant to the solution

= - e^{- x} sin (e^{- x}) - cos (e^{- x}) + C

\frac{d}{d y} (\frac{x + y}{x - y})

\frac{d}{d x} (\frac{e ^{2 x + 1}}{( x - 1 ) ^{2}})

\int \frac{1}{2} sin (2 x) d x

F (t) = 2 + 4 e^{- 4 t} - 3 e^{3 t}

\frac{\partial}{\partial x} (ln (x + \frac{y}{2 x}))

What is the integral of e^{-2x}cos(e^{-x}) ?
The integral of e^{-2x}cos(e^{-x}) is -e^{-x}sin(e^{-x})-cos(e^{-x})+C