What is the integral of 2sin(3x)e^{-x} ?

The integral of 2sin(3x)e^{-x} is 2(-(3e^{-x}cos(3x))/(10)-(e^{-x}sin(3x))/(10))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 2sin(3x)e^{-x}

\int 2 sin (3 x) e^{- x} d x

2 (- \frac{3 e ^{- x} cos ( 3 x )}{10} - \frac{e ^{- x} sin ( 3 x )}{10}) + C

Solution steps

\int 2 sin (3 x) e^{- x} d x

Take the constant out:

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

= 2 \cdot \int sin (3 x) e^{- x} d x

Apply Integration By Parts

: - \frac{1}{3} e^{- x} cos (3 x) - \int \frac{1}{3} e^{- x} cos (3 x) d x

= 2 (- \frac{1}{3} e^{- x} cos (3 x) - \int \frac{1}{3} e^{- x} cos (3 x) d x)

Take the constant out:

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

= 2 (- \frac{1}{3} e^{- x} cos (3 x) - \frac{1}{3} \cdot \int e^{- x} cos (3 x) d x)

Apply Integration By Parts

: \frac{1}{3} e^{- x} sin (3 x) - \int - \frac{1}{3} e^{- x} sin (3 x) d x

= 2 (- \frac{1}{3} e^{- x} cos (3 x) - \frac{1}{3} (\frac{1}{3} e^{- x} sin (3 x) - \int - \frac{1}{3} e^{- x} sin (3 x) d x))

Take the constant out:

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

= 2 (- \frac{1}{3} e^{- x} cos (3 x) - \frac{1}{3} (\frac{1}{3} e^{- x} sin (3 x) - (- \frac{1}{3} \cdot \int e^{- x} sin (3 x) d x)))

Therefore

2 \cdot \int e^{- x} sin (3 x) d x = 2 (- \frac{1}{3} e^{- x} cos (3 x) - \frac{1}{3} (\frac{1}{3} e^{- x} sin (3 x) - (- \frac{1}{3} \cdot \int e^{- x} sin (3 x) d x)))

Isolate

\int e^{- x} sin (3 x) d x

= 2 (- \frac{3 e ^{- x} cos ( 3 x )}{10} - \frac{e ^{- x} sin ( 3 x )}{10})

Add a constant to the solution

= 2 (- \frac{3 e ^{- x} cos ( 3 x )}{10} - \frac{e ^{- x} sin ( 3 x )}{10}) + C

\int 2^{7 x + 3} d x

\int cos (5 x) cos (6 x) d x

\int (8 x^{3} - 4 x) (x^{4} - x^{2})^{3} d x

\int 9 x sin (x) d x

\int e^{- θ} cos (\frac{θ}{2}) d θ

What is the integral of 2sin(3x)e^{-x} ?
The integral of 2sin(3x)e^{-x} is 2(-(3e^{-x}cos(3x))/(10)-(e^{-x}sin(3x))/(10))+C