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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

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\int_{0}^{1} π (x^{4} - x) d x

\frac{\partial}{\partial x} (e^{e^{y} x + y})

f (x) = 4 x^{2} + 6 x + 3

\int e^{- (x)^{2}} d x

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