What is the integral of 4e^{5t}sin(3t) ?

The integral of 4e^{5t}sin(3t) is 4(-(3e^{5t}cos(3t))/(34)+(5e^{5t}sin(3t))/(34))+C

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integral of 4e^{5t}sin(3t)

\int 4 e^{5 t} sin (3 t) d t

4 (- \frac{3 e ^{5 t} cos ( 3 t )}{34} + \frac{5 e ^{5 t} sin ( 3 t )}{34}) + C

Solution steps

\int 4 e^{5 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 \cdot \int e^{5 t} sin (3 t) d t

Apply Integration By Parts

= 4 (- \frac{1}{3} e^{5 t} cos (3 t) - \int - \frac{5}{3} e^{5 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^{5 t} cos (3 t) - (- \frac{5}{3} \cdot \int e^{5 t} cos (3 t) d t))

Apply Integration By Parts

= 4 (- \frac{1}{3} e^{5 t} cos (3 t) - (- \frac{5}{3} (\frac{1}{3} e^{5 t} sin (3 t) - \int \frac{5}{3} e^{5 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^{5 t} cos (3 t) - (- \frac{5}{3} (\frac{1}{3} e^{5 t} sin (3 t) - \frac{5}{3} \cdot \int e^{5 t} sin (3 t) d t)))

Therefore

4 \cdot \int e^{5 t} sin (3 t) d t = 4 (- \frac{1}{3} e^{5 t} cos (3 t) - (- \frac{5}{3} (\frac{1}{3} e^{5 t} sin (3 t) - \frac{5}{3} \cdot \int e^{5 t} sin (3 t) d t)))

Isolate

\int e^{5 t} sin (3 t) d t

= 4 (- \frac{3 e ^{5 t} cos ( 3 t )}{34} + \frac{5 e ^{5 t} sin ( 3 t )}{34})

Add a constant to the solution

= 4 (- \frac{3 e ^{5 t} cos ( 3 t )}{34} + \frac{5 e ^{5 t} sin ( 3 t )}{34}) + C

What is the integral of 4e^{5t}sin(3t) ?
The integral of 4e^{5t}sin(3t) is 4(-(3e^{5t}cos(3t))/(34)+(5e^{5t}sin(3t))/(34))+C