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

The integral of e^{-3t}cos(t) is (e^{-3t}sin(t))/(10)-(3e^{-3t}cos(t))/(10)+C

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

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

\frac{e ^{- 3 t} sin ( t )}{10} - \frac{3 e ^{- 3 t} cos ( t )}{10} + C

Solution steps

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

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

= \frac{e ^{- 3 t} sin ( t )}{10} - \frac{3 e ^{- 3 t} cos ( t )}{10}

Add a constant to the solution

= \frac{e ^{- 3 t} sin ( t )}{10} - \frac{3 e ^{- 3 t} cos ( t )}{10} + C

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What is the integral of e^{-3t}cos(t) ?
The integral of e^{-3t}cos(t) is (e^{-3t}sin(t))/(10)-(3e^{-3t}cos(t))/(10)+C