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

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

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integral of e^{2t}sin(t)

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

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

Solution steps

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

Apply Integration By Parts

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

Take the constant out:

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

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

Apply Integration By Parts

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

Take the constant out:

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

d er i v a t i v e a^{6} + cos^{6} (x)

\int_{- 3}^{1} (1 + 4 x) d x

x \to 6 lim (2 x - 5)

\frac{d x}{d t} = 3 x t

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

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