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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

= 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} sin (t) - 2 \cdot \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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{1}{4} x^{2})

h \to 0 lim (sin (h))

\frac{d}{d y} (2 x y^{2})

(e^{x} - 1) sin (- 2 x)

\int e^{7 x} - \frac{1}{x} d x

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