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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

k = 1 \sum \infty \frac{4 k}{7 k - 1}

5 cos (2 t)

\int_{1}^{0} x^{2} d x

\frac{d}{d x} (sin^{3} (2 x) + x^{2})

y^{^{''}} - 4 y + 3 y = 0

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