What is the integral of e^{4t}cos(t) ?

The integral of e^{4t}cos(t) is (e^{4t}sin(t))/(17)+(4e^{4t}cos(t))/(17)+C

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integral of e^{4t}cos(t)

\int e^{4 t} cos (t) d t

\frac{e ^{4 t} sin ( t )}{17} + \frac{4 e ^{4 t} cos ( t )}{17} + C

Solution steps

\int e^{4 t} cos (t) d t

Apply Integration By Parts

: e^{4 t} sin (t) - \int e^{4 t} \cdot 4 sin (t) d t

= e^{4 t} sin (t) - \int e^{4 t} \cdot 4 sin (t) d t

Take the constant out:

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

= e^{4 t} sin (t) - 4 \cdot \int e^{4 t} sin (t) d t

Apply Integration By Parts

: - e^{4 t} cos (t) - \int - 4 e^{4 t} cos (t) d t

= e^{4 t} sin (t) - 4 (- e^{4 t} cos (t) - \int - 4 e^{4 t} cos (t) d t)

Take the constant out:

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

= e^{4 t} sin (t) - 4 (- e^{4 t} cos (t) - (- 4 \cdot \int e^{4 t} cos (t) d t))

Therefore

\int e^{4 t} cos (t) d t = e^{4 t} sin (t) - 4 (- e^{4 t} cos (t) - (- 4 \cdot \int e^{4 t} cos (t) d t))

Isolate

\int e^{4 t} cos (t) d t

= \frac{e ^{4 t} sin ( t )}{17} + \frac{4 e ^{4 t} cos ( t )}{17}

Add a constant to the solution

= \frac{e ^{4 t} sin ( t )}{17} + \frac{4 e ^{4 t} cos ( t )}{17} + C

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What is the integral of e^{4t}cos(t) ?
The integral of e^{4t}cos(t) is (e^{4t}sin(t))/(17)+(4e^{4t}cos(t))/(17)+C