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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Simplify

- a \frac{1}{b} : - \frac{a}{b}

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Simplify

a \frac{1}{b} : \frac{a}{b}

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (\frac{x}{y + z})

\int_{0}^{1} \frac{3}{2 x - 5} d x

\int_{0}^{1} 19 \frac{1}{x ^{p}} d x

f (x) = 7 x^{6} e^{x} + e^{x} x^{7}

\frac{3}{s ( 6 s ^{2} + 5 s + 1 )}

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