What is the integral of e^tsin(3t) ?

The integral of e^tsin(3t) is -(3e^tcos(3t))/(10)+(e^tsin(3t))/(10)+C

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integral of e^tsin(3t)

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

- \frac{3 e ^{t} cos ( 3 t )}{10} + \frac{e ^{t} sin ( 3 t )}{10} + C

Solution steps

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

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

= - \frac{3 e ^{t} cos ( 3 t )}{10} + \frac{e ^{t} sin ( 3 t )}{10}

Add a constant to the solution

= - \frac{3 e ^{t} cos ( 3 t )}{10} + \frac{e ^{t} sin ( 3 t )}{10} + C

2 tan^{2} (4 x)

\frac{\partial}{\partial x} (x^{2} y)

y = lo g_{3} (sin (x^{2}))

y^{^{'}} + 2 y = x y^{- 2}

\frac{d}{d x} ((x - 1)^{2 x})

What is the integral of e^tsin(3t) ?
The integral of e^tsin(3t) is -(3e^tcos(3t))/(10)+(e^tsin(3t))/(10)+C