What is the integral of e^{-2t}cos(3t) ?

The integral of e^{-2t}cos(3t) is (3e^{-2t}sin(3t))/(13)-(2e^{-2t}cos(3t))/(13)+C

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integral of e^{-2t}cos(3t)

\int e^{- 2 t} cos (3 t) d t

\frac{3 e ^{- 2 t} sin ( 3 t )}{13} - \frac{2 e ^{- 2 t} cos ( 3 t )}{13} + C

Solution steps

\int e^{- 2 t} cos (3 t) d t

Apply Integration By Parts

= \frac{1}{3} e^{- 2 t} sin (3 t) - \int - \frac{2}{3} e^{- 2 t} sin (3 t) d t

Take the constant out:

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

= \frac{1}{3} e^{- 2 t} sin (3 t) - (- \frac{2}{3} \cdot \int e^{- 2 t} sin (3 t) d t)

Apply Integration By Parts

= \frac{1}{3} e^{- 2 t} sin (3 t) - (- \frac{2}{3} (- \frac{1}{3} e^{- 2 t} cos (3 t) - \int \frac{2}{3} e^{- 2 t} cos (3 t) d t))

Take the constant out:

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

= \frac{1}{3} e^{- 2 t} sin (3 t) - (- \frac{2}{3} (- \frac{1}{3} e^{- 2 t} cos (3 t) - \frac{2}{3} \cdot \int e^{- 2 t} cos (3 t) d t))

Therefore

\int e^{- 2 t} cos (3 t) d t = \frac{1}{3} e^{- 2 t} sin (3 t) - (- \frac{2}{3} (- \frac{1}{3} e^{- 2 t} cos (3 t) - \frac{2}{3} \cdot \int e^{- 2 t} cos (3 t) d t))

Isolate

\int e^{- 2 t} cos (3 t) d t

= \frac{3 e ^{- 2 t} sin ( 3 t )}{13} - \frac{2 e ^{- 2 t} cos ( 3 t )}{13}

Add a constant to the solution

= \frac{3 e ^{- 2 t} sin ( 3 t )}{13} - \frac{2 e ^{- 2 t} cos ( 3 t )}{13} + C

What is the integral of e^{-2t}cos(3t) ?
The integral of e^{-2t}cos(3t) is (3e^{-2t}sin(3t))/(13)-(2e^{-2t}cos(3t))/(13)+C