What is the integral of e^{2x}cos(3x+1) ?

The integral of e^{2x}cos(3x+1) is (3e^{2x}sin(3x+1))/(13)+(2e^{2x}cos(3x+1))/(13)+C

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integral of e^{2x}cos(3x+1)

\int e^{2 x} cos (3 x + 1) d x

\frac{3 e ^{2 x} sin ( 3 x + 1 )}{13} + \frac{2 e ^{2 x} cos ( 3 x + 1 )}{13} + C

Solution steps

\int e^{2 x} cos (3 x + 1) d x

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

\int e^{2 x} cos (3 x + 1) d x

= \frac{3 e ^{2 x} sin ( 3 x + 1 )}{13} + \frac{2 e ^{2 x} cos ( 3 x + 1 )}{13}

Add a constant to the solution

= \frac{3 e ^{2 x} sin ( 3 x + 1 )}{13} + \frac{2 e ^{2 x} cos ( 3 x + 1 )}{13} + C

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What is the integral of e^{2x}cos(3x+1) ?
The integral of e^{2x}cos(3x+1) is (3e^{2x}sin(3x+1))/(13)+(2e^{2x}cos(3x+1))/(13)+C