What is the integral of (cos(2x)e^{5x})/2 ?

The integral of (cos(2x)e^{5x})/2 is 1/58 (2e^{5x}sin(2x)+5e^{5x}cos(2x))+C

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integral of (cos(2x)e^{5x})/2

\int \frac{cos ( 2 x ) e ^{5 x}}{2} d x

\frac{1}{58} (2 e^{5 x} sin (2 x) + 5 e^{5 x} cos (2 x)) + C

Solution steps

\int \frac{cos ( 2 x ) e ^{5 x}}{2} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int cos (2 x) e^{5 x} d x

Apply Integration By Parts

: \frac{1}{2} e^{5 x} sin (2 x) - \int \frac{5}{2} e^{5 x} sin (2 x) d x

= \frac{1}{2} (\frac{1}{2} e^{5 x} sin (2 x) - \int \frac{5}{2} e^{5 x} sin (2 x) d x)

Take the constant out:

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

= \frac{1}{2} (\frac{1}{2} e^{5 x} sin (2 x) - \frac{5}{2} \cdot \int e^{5 x} sin (2 x) d x)

Apply Integration By Parts

: - \frac{1}{2} e^{5 x} cos (2 x) - \int - \frac{5}{2} e^{5 x} cos (2 x) d x

= \frac{1}{2} (\frac{1}{2} e^{5 x} sin (2 x) - \frac{5}{2} (- \frac{1}{2} e^{5 x} cos (2 x) - \int - \frac{5}{2} e^{5 x} cos (2 x) d x))

Take the constant out:

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

= \frac{1}{2} (\frac{1}{2} e^{5 x} sin (2 x) - \frac{5}{2} (- \frac{1}{2} e^{5 x} cos (2 x) - (- \frac{5}{2} \cdot \int e^{5 x} cos (2 x) d x)))

Therefore

\frac{1}{2} \cdot \int e^{5 x} cos (2 x) d x = \frac{1}{2} (\frac{1}{2} e^{5 x} sin (2 x) - \frac{5}{2} (- \frac{1}{2} e^{5 x} cos (2 x) - (- \frac{5}{2} \cdot \int e^{5 x} cos (2 x) d x)))

Isolate

\int e^{5 x} cos (2 x) d x

= \frac{1}{2} (\frac{2 e ^{5 x} sin ( 2 x )}{29} + \frac{5 e ^{5 x} cos ( 2 x )}{29})

Simplify

\frac{1}{2} (\frac{2 e ^{5 x} sin ( 2 x )}{29} + \frac{5 e ^{5 x} cos ( 2 x )}{29}) : \frac{1}{58} (2 e^{5 x} sin (2 x) + 5 e^{5 x} cos (2 x))

= \frac{1}{58} (2 e^{5 x} sin (2 x) + 5 e^{5 x} cos (2 x))

Add a constant to the solution

= \frac{1}{58} (2 e^{5 x} sin (2 x) + 5 e^{5 x} cos (2 x)) + C

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What is the integral of (cos(2x)e^{5x})/2 ?
The integral of (cos(2x)e^{5x})/2 is 1/58 (2e^{5x}sin(2x)+5e^{5x}cos(2x))+C