What is the integral of 4cos(3t)(e^{2t}) ?

The integral of 4cos(3t)(e^{2t}) is 4((3e^{2t}sin(3t))/(13)+(2e^{2t}cos(3t))/(13))+C

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integral of 4cos(3t)(e^{2t})

\int 4 cos (3 t) (e^{2 t}) d t

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

Solution steps

\int 4 cos (3 t) e^{2 t} d t

Take the constant out:

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

= 4 \cdot \int cos (3 t) e^{2 t} d t

Apply Integration By Parts

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

= 4 (\frac{1}{3} e^{2 t} sin (3 t) - \int \frac{2}{3} e^{2 t} sin (3 t) d t)

Take the constant out:

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

= 4 (\frac{1}{3} e^{2 t} sin (3 t) - \frac{2}{3} \cdot \int e^{2 t} sin (3 t) d t)

Apply Integration By Parts

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

= 4 (\frac{1}{3} e^{2 t} sin (3 t) - \frac{2}{3} (- \frac{1}{3} e^{2 t} cos (3 t) - \int - \frac{2}{3} e^{2 t} cos (3 t) d t))

Take the constant out:

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

= 4 (\frac{1}{3} e^{2 t} sin (3 t) - \frac{2}{3} (- \frac{1}{3} e^{2 t} cos (3 t) - (- \frac{2}{3} \cdot \int e^{2 t} cos (3 t) d t)))

Therefore

4 \cdot \int e^{2 t} cos (3 t) d t = 4 (\frac{1}{3} e^{2 t} sin (3 t) - \frac{2}{3} (- \frac{1}{3} e^{2 t} cos (3 t) - (- \frac{2}{3} \cdot \int e^{2 t} cos (3 t) d t)))

Isolate

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

= 4 (\frac{3 e ^{2 t} sin ( 3 t )}{13} + \frac{2 e ^{2 t} cos ( 3 t )}{13})

Add a constant to the solution

= 4 (\frac{3 e ^{2 t} sin ( 3 t )}{13} + \frac{2 e ^{2 t} cos ( 3 t )}{13}) + C

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What is the integral of 4cos(3t)(e^{2t}) ?
The integral of 4cos(3t)(e^{2t}) is 4((3e^{2t}sin(3t))/(13)+(2e^{2t}cos(3t))/(13))+C