What is the integral of e^xcos(ax) ?

The integral of e^xcos(ax) is (ae^xsin(ax))/(a^2+1)+(e^xcos(ax))/(a^2+1)+C

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integral of e^xcos(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

x \to + 4 lim (\frac{2}{( x + 4 ) ^{3}})

x \to 0 - lim (\frac{x}{x})

\frac{d}{d x} (\frac{x ^{2} - 3 x + 2}{x ^{7} - 2})

\frac{d y}{d x} + \frac{2 x}{1 + x ^{2}} y = \frac{1}{1 + x ^{2}}

\int \frac{x ^{2} + 3 x - 2}{x ^{2} - 1} d x

What is the integral of e^xcos(ax) ?
The integral of e^xcos(ax) is (ae^xsin(ax))/(a^2+1)+(e^xcos(ax))/(a^2+1)+C