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

The integral of e^{2x}*sin(-3x) is -2/39 (2e^{2x}cos(3x)+3e^{2x}sin(3x))+1/3 e^{2x}cos(3x)+C

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

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

- \frac{2}{39} (2 e^{2 x} cos (3 x) + 3 e^{2 x} sin (3 x)) + \frac{1}{3} e^{2 x} cos (3 x) + C

Solution steps

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

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

= - \frac{2}{3} (\frac{2 e ^{2 x} cos ( 3 x )}{13} + \frac{3 e ^{2 x} sin ( 3 x )}{13}) + cos (3 x) \frac{1}{3} e^{2 x}

Simplify

- \frac{2}{3} (\frac{2 e ^{2 x} cos ( 3 x )}{13} + \frac{3 e ^{2 x} sin ( 3 x )}{13}) + cos (3 x) \frac{1}{3} e^{2 x} : - \frac{2}{39} (2 e^{2 x} cos (3 x) + 3 e^{2 x} sin (3 x)) + \frac{1}{3} e^{2 x} cos (3 x)

= - \frac{2}{39} (2 e^{2 x} cos (3 x) + 3 e^{2 x} sin (3 x)) + \frac{1}{3} e^{2 x} cos (3 x)

Add a constant to the solution

= - \frac{2}{39} (2 e^{2 x} cos (3 x) + 3 e^{2 x} sin (3 x)) + \frac{1}{3} e^{2 x} cos (3 x) + C

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f (x) = x^{2} - 3, at x = 2

n = 2 \sum \infty \frac{2}{5 ^{n}}

\int_{- 6}^{- 1} 2 x^{\frac{1}{3}} d x

What is the integral of e^{2x}*sin(-3x) ?
The integral of e^{2x}*sin(-3x) is -2/39 (2e^{2x}cos(3x)+3e^{2x}sin(3x))+1/3 e^{2x}cos(3x)+C