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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

(x y^{3} + 1) d x + x^{2} y^{2} d y = 0

x \frac{d y}{d x} - x^{2} y = - x^{2}

\frac{y ^{^{'}} - e ^{- t} + 4}{y} = - 4, y (0) = 3

o f (o, y) y = \frac{π}{2}, (π, \frac{π}{2})

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

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