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

The integral of e^x(cos(x)) is (e^xsin(x))/2+(e^xcos(x))/2+C

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integral of e^x(cos(x))

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

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

Solution steps

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

Apply Integration By Parts

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

= e^{x} sin (x) - \int e^{x} sin (x) d x

Apply Integration By Parts

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

= e^{x} sin (x) - (- e^{x} cos (x) - \int - e^{x} cos (x) d x)

Take the constant out:

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

= e^{x} sin (x) - (- e^{x} cos (x) - (- \int e^{x} cos (x) d x))

Therefore

\int e^{x} cos (x) d x = e^{x} sin (x) - (- e^{x} cos (x) - (- \int e^{x} cos (x) d x))

Isolate

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

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

Add a constant to the solution

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

x + 2, x^{2}

\int x \cdot 4 - x^{2} d x

x^{2} y^{^{''}} - 2 x y^{^{'}} + (x^{2} + 2) y = 0

y = \frac{x}{5} - \frac{5}{x}

x \to 1 lim (\frac{x}{ln ( 1 + x )})

What is the integral of e^x(cos(x)) ?
The integral of e^x(cos(x)) is (e^xsin(x))/2+(e^xcos(x))/2+C