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

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

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

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

- e^{- 2 x} sin (2 y) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int - \frac{1}{2} e^{u} d u

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

Take the constant out:

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

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

Use the common integral:

\int e^{u} d u = e^{u}

= - 2 sin (- 2 y) (- \frac{1}{2} e^{u})

Substitute back

u = - 2 x

= - 2 sin (- 2 y) (- \frac{1}{2} e^{- 2 x})

Simplify

- 2 sin (- 2 y) (- \frac{1}{2} e^{- 2 x}) : - e^{- 2 x} sin (2 y)

= - e^{- 2 x} sin (2 y)

Add a constant to the solution

= - e^{- 2 x} sin (2 y) + C

f (x) = x^{4} + \frac{2}{x}

n = 0 \sum \infty \frac{n}{n ^{2} + 9}

f (x) = \frac{x ^{2} + 4}{x}

\frac{d}{d x} (\frac{x ^{3} - x ^{2} + 1}{x - 2})

a x^{2}

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