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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

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