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

The integral of e^{-sx}cos(x) is (e^{-sx}sin(x))/(1+s^2)-(se^{-sx}cos(x))/(1+s^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{e ^{- s x} sin ( x )}{1 + s ^{2}} - \frac{s e ^{- s x} cos ( x )}{1 + s ^{2}} + C

Solution steps

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

Apply Integration By Parts

: e^{- s x} sin (x) - \int - s e^{- s x} sin (x) d x

= e^{- s x} sin (x) - \int - s e^{- s x} sin (x) d x

Take the constant out:

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

= e^{- s x} sin (x) - (- s \cdot \int e^{- s x} sin (x) d x)

Apply Integration By Parts

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

= e^{- s x} sin (x) - (- s (- e^{- s x} cos (x) - \int s e^{- s x} cos (x) d x))

Take the constant out:

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

= e^{- s x} sin (x) - (- s (- e^{- s x} cos (x) - s \cdot \int e^{- s x} cos (x) d x))

Therefore

\int e^{- s x} cos (x) d x = e^{- s x} sin (x) - (- s (- e^{- s x} cos (x) - s \cdot \int e^{- s x} cos (x) d x))

Isolate

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

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

Add a constant to the solution

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

4 \frac{d y}{d t} + y = 36 t, y (0) = 6

\int_{- \infty}^{2} e^{- x} d x

\frac{sin ( x )}{1 + cos ( x )}

x \to 2 lim (\frac{x}{x ^{2} - 4})

(5 x^{2} + 4) (3 x - 2)

What is the integral of e^{-sx}cos(x) ?
The integral of e^{-sx}cos(x) is (e^{-sx}sin(x))/(1+s^2)-(se^{-sx}cos(x))/(1+s^2)+C