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

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

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integral of e^tcos(2t)

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

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

Solution steps

\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

= 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

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

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

\int \frac{x ^{3 n}}{n !} d x

\frac{1}{s ^{2} + 2 s + 4}

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

f (x) = (2 x^{3} + 5)^{10}

\frac{d}{d x} (e^{l n (2 x)})

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