What is the integral of e^{3x}cos(2x) ?

The integral of e^{3x}cos(2x) is (2e^{3x}sin(2x))/(13)+(3e^{3x}cos(2x))/(13)+C

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integral of e^{3x}cos(2x)

\int e^{3 x} cos (2 x) d x

\frac{2 e ^{3 x} sin ( 2 x )}{13} + \frac{3 e ^{3 x} cos ( 2 x )}{13} + C

Solution steps

\int e^{3 x} cos (2 x) d x

Apply Integration By Parts

: \frac{1}{2} e^{3 x} sin (2 x) - \int \frac{3}{2} e^{3 x} sin (2 x) d x

= \frac{1}{2} e^{3 x} sin (2 x) - \int \frac{3}{2} e^{3 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} e^{3 x} sin (2 x) - \frac{3}{2} \cdot \int e^{3 x} sin (2 x) d x

Apply Integration By Parts

: - \frac{1}{2} e^{3 x} cos (2 x) - \int - \frac{3}{2} e^{3 x} cos (2 x) d x

= \frac{1}{2} e^{3 x} sin (2 x) - \frac{3}{2} (- \frac{1}{2} e^{3 x} cos (2 x) - \int - \frac{3}{2} e^{3 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} e^{3 x} sin (2 x) - \frac{3}{2} (- \frac{1}{2} e^{3 x} cos (2 x) - (- \frac{3}{2} \cdot \int e^{3 x} cos (2 x) d x))

Therefore

\int e^{3 x} cos (2 x) d x = \frac{1}{2} e^{3 x} sin (2 x) - \frac{3}{2} (- \frac{1}{2} e^{3 x} cos (2 x) - (- \frac{3}{2} \cdot \int e^{3 x} cos (2 x) d x))

Isolate

\int e^{3 x} cos (2 x) d x

= \frac{2 e ^{3 x} sin ( 2 x )}{13} + \frac{3 e ^{3 x} cos ( 2 x )}{13}

Add a constant to the solution

= \frac{2 e ^{3 x} sin ( 2 x )}{13} + \frac{3 e ^{3 x} cos ( 2 x )}{13} + C

5 x^{2} cos (x) + 4 x

y^{^{''}} - 2 y^{^{'}} - 8 y = 0, y^{^{'}} (0) = 4, y (0) = - 2

x \to 0 lim (ln (tan (2 x)))

\int sin (π) x d x

f (x) = x^{2} + 2, (1, 3)

What is the integral of e^{3x}cos(2x) ?
The integral of e^{3x}cos(2x) is (2e^{3x}sin(2x))/(13)+(3e^{3x}cos(2x))/(13)+C