What is the integral of (e^x)*cos(nx) ?

The integral of (e^x)*cos(nx) is (ne^xsin(nx))/(n^2+1)+(e^xcos(nx))/(n^2+1)+C

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integral of (e^x)*cos(nx)

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

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

Solution steps

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

Apply Integration By Parts

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

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

Apply Integration By Parts

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

\int x x - 10 d x

\frac{3}{s} + \frac{16}{s ( s + 1 )}

\int_{1}^{e^{2}} \frac{ln ( x )}{x ^{2}} d x

\frac{\partial}{\partial x} (\frac{a - b ^{2}}{x - c ^{2}})

\frac{d}{d x} (5 - \frac{4}{x})

What is the integral of (e^x)*cos(nx) ?
The integral of (e^x)*cos(nx) is (ne^xsin(nx))/(n^2+1)+(e^xcos(nx))/(n^2+1)+C