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

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

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

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

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

Solution steps

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

Apply Integration By Parts

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

Take the constant out:

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

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

Apply Integration By Parts

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

Take the constant out:

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

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

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