What is the integral of e^tsin(2t) ?

The integral of e^tsin(2t) is -(2e^tcos(2t))/5+(e^tsin(2t))/5+C

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integral of e^tsin(2t)

\int e^{t} sin (2 t) d t

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

Solution steps

\int e^{t} sin (2 t) d t

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

\int e^{t} sin (2 t) d t

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

Add a constant to the solution

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

y = 6 + 5 x^{2} - 2 x^{3}

\frac{d}{d x} (cos^{7} (2 x^{2} + 1))

\frac{d}{d x} ((\frac{x + 3}{x - 3})^{5})

x \to \infty lim (2 - x)

f (x) = x^{2} e^{\frac{1}{x}}

What is the integral of e^tsin(2t) ?
The integral of e^tsin(2t) is -(2e^tcos(2t))/5+(e^tsin(2t))/5+C