What is the integral of 2e^{-x}cos(5x) ?

The integral of 2e^{-x}cos(5x) is 2((5e^{-x}sin(5x))/(26)-(e^{-x}cos(5x))/(26))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 2e^{-x}cos(5x)

\int 2 e^{- x} cos (5 x) d x

2 (\frac{5 e ^{- x} sin ( 5 x )}{26} - \frac{e ^{- x} cos ( 5 x )}{26}) + C

Solution steps

\int 2 e^{- x} cos (5 x) d x

Take the constant out:

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

= 2 \cdot \int e^{- x} cos (5 x) d x

Apply Integration By Parts

: \frac{1}{5} e^{- x} sin (5 x) - \int - \frac{1}{5} e^{- x} sin (5 x) d x

= 2 (\frac{1}{5} e^{- x} sin (5 x) - \int - \frac{1}{5} e^{- x} sin (5 x) d x)

Take the constant out:

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

= 2 (\frac{1}{5} e^{- x} sin (5 x) - (- \frac{1}{5} \cdot \int e^{- x} sin (5 x) d x))

Apply Integration By Parts

: - \frac{1}{5} e^{- x} cos (5 x) - \int \frac{1}{5} e^{- x} cos (5 x) d x

= 2 (\frac{1}{5} e^{- x} sin (5 x) - (- \frac{1}{5} (- \frac{1}{5} e^{- x} cos (5 x) - \int \frac{1}{5} e^{- x} cos (5 x) d x)))

Take the constant out:

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

= 2 (\frac{1}{5} e^{- x} sin (5 x) - (- \frac{1}{5} (- \frac{1}{5} e^{- x} cos (5 x) - \frac{1}{5} \cdot \int e^{- x} cos (5 x) d x)))

Therefore

2 \cdot \int e^{- x} cos (5 x) d x = 2 (\frac{1}{5} e^{- x} sin (5 x) - (- \frac{1}{5} (- \frac{1}{5} e^{- x} cos (5 x) - \frac{1}{5} \cdot \int e^{- x} cos (5 x) d x)))

Isolate

\int e^{- x} cos (5 x) d x

= 2 (\frac{5 e ^{- x} sin ( 5 x )}{26} - \frac{e ^{- x} cos ( 5 x )}{26})

Add a constant to the solution

= 2 (\frac{5 e ^{- x} sin ( 5 x )}{26} - \frac{e ^{- x} cos ( 5 x )}{26}) + C

\int \frac{1}{5 3 x ^{5}} d x

\int x^{3} x^{4} + 7 d x

\int \frac{x ^{3} + 2 x}{3 x ^{2} + 5} d x

\int \frac{x ^{\frac{1}{2}}}{x + x ^{\frac{1}{4}}} d x

\int \frac{5 ^{x}}{6 ^{x}} d x

What is the integral of 2e^{-x}cos(5x) ?
The integral of 2e^{-x}cos(5x) is 2((5e^{-x}sin(5x))/(26)-(e^{-x}cos(5x))/(26))+C