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

The integral of e^{ax}cos(nx) is (ne^{ax}sin(nx))/(n^2+a^2)+(ae^{ax}cos(nx))/(n^2+a^2)+C

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integral of e^{ax}cos(nx)

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

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

Solution steps

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

Apply Integration By Parts

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

Simplify

a \frac{1}{n} : \frac{a}{n}

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

Apply Integration By Parts

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

Simplify

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

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What is the integral of e^{ax}cos(nx) ?
The integral of e^{ax}cos(nx) is (ne^{ax}sin(nx))/(n^2+a^2)+(ae^{ax}cos(nx))/(n^2+a^2)+C