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

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

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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