What is the integral of e^xcos(5x) ?

The integral of e^xcos(5x) is (5e^xsin(5x))/(26)+(e^xcos(5x))/(26)+C

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integral of e^xcos(5x)

\int e^{x} cos (5 x) d x

\frac{5 e ^{x} sin ( 5 x )}{26} + \frac{e ^{x} cos ( 5 x )}{26} + C

Solution steps

\int e^{x} cos (5 x) d x

Apply Integration By Parts

: e^{x} \frac{1}{5} sin (5 x) - \int e^{x} \frac{1}{5} sin (5 x) d x

= e^{x} \frac{1}{5} sin (5 x) - \int e^{x} \frac{1}{5} sin (5 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}{5} sin (5 x) - \frac{1}{5} \cdot \int e^{x} sin (5 x) d x

Apply Integration By Parts

: - \frac{1}{5} e^{x} cos (5 x) - \int - \frac{1}{5} e^{x} cos (5 x) d x

= e^{x} \frac{1}{5} sin (5 x) - \frac{1}{5} (- \frac{1}{5} e^{x} cos (5 x) - \int - \frac{1}{5} e^{x} cos (5 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}{5} sin (5 x) - \frac{1}{5} (- \frac{1}{5} e^{x} cos (5 x) - (- \frac{1}{5} \cdot \int e^{x} cos (5 x) d x))

Therefore

\int e^{x} cos (5 x) d x = e^{x} \frac{1}{5} sin (5 x) - \frac{1}{5} (- \frac{1}{5} e^{x} cos (5 x) - (- \frac{1}{5} \cdot \int e^{x} cos (5 x) d x))

Isolate

\int e^{x} cos (5 x) d x

= \frac{5 e ^{x} sin ( 5 x )}{26} + \frac{e ^{x} cos ( 5 x )}{26}

Add a constant to the solution

= \frac{5 e ^{x} sin ( 5 x )}{26} + \frac{e ^{x} cos ( 5 x )}{26} + C

- 60 = \frac{d ^{2} y}{d x ^{2}}

f (x) = 4 e^{0.5 x}

\frac{\partial}{\partial x} (x^{5} + x^{2} y + x + 5)

y^{^{''}} - 4 y^{^{'}} + 3 y = 0, y (0) = 1, y^{^{'}} (0) = 3

x \to 1 lim (e^{x^{2} - 4 x + 1})

What is the integral of e^xcos(5x) ?
The integral of e^xcos(5x) is (5e^xsin(5x))/(26)+(e^xcos(5x))/(26)+C