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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

Take the constant out:

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

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

Apply Integration By Parts

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

Take the constant out:

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

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