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

The integral of e^{-x}cos(ax) is (ae^{-x}sin(ax))/(a^2+1)-(e^{-x}cos(ax))/(a^2+1)+C

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integral of e^{-x}cos(ax)

\int e^{- x} cos (a x) d x

\frac{a e ^{- x} sin ( a x )}{a ^{2} + 1} - \frac{e ^{- x} cos ( a x )}{a ^{2} + 1} + C

Solution steps

\int e^{- x} cos (a x) d x

Apply Integration By Parts

: e^{- x} \frac{1}{a} sin (a x) - \int - e^{- x} \frac{1}{a} sin (a x) d x

= e^{- x} \frac{1}{a} sin (a x) - \int - e^{- x} \frac{1}{a} sin (a x) d x

Take the constant out:

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

= e^{- x} \frac{1}{a} sin (a x) - (- \frac{1}{a} \cdot \int e^{- x} sin (a x) d x)

Apply Integration By Parts

: - e^{- x} \frac{1}{a} cos (a x) - \int e^{- x} \frac{1}{a} cos (a x) d x

= e^{- x} \frac{1}{a} sin (a x) - (- \frac{1}{a} (- e^{- x} \frac{1}{a} cos (a x) - \int e^{- x} \frac{1}{a} cos (a x) d x))

Take the constant out:

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

= e^{- x} \frac{1}{a} sin (a x) - (- \frac{1}{a} (- e^{- x} \frac{1}{a} cos (a x) - \frac{1}{a} \cdot \int e^{- x} cos (a x) d x))

Therefore

\int e^{- x} cos (a x) d x = e^{- x} \frac{1}{a} sin (a x) - (- \frac{1}{a} (- e^{- x} \frac{1}{a} cos (a x) - \frac{1}{a} \cdot \int e^{- x} cos (a x) d x))

Isolate

\int e^{- x} cos (a x) d x

= \frac{a e ^{- x} sin ( a x )}{a ^{2} + 1} - \frac{e ^{- x} cos ( a x )}{a ^{2} + 1}

Add a constant to the solution

= \frac{a e ^{- x} sin ( a x )}{a ^{2} + 1} - \frac{e ^{- x} cos ( a x )}{a ^{2} + 1} + C

\int 2 x e^{- y} d x

\frac{d}{d x} (4 cos (x) + 2 sin (x))

x \to \infty lim (π^{x})

f (x) = \frac{2}{x + 1}, (4, \frac{2}{5})

2 \cdot e^{3 x}

What is the integral of e^{-x}cos(ax) ?
The integral of e^{-x}cos(ax) is (ae^{-x}sin(ax))/(a^2+1)-(e^{-x}cos(ax))/(a^2+1)+C