What is the integral of e^{3x}sin(3x) ?

The integral of e^{3x}sin(3x) is -(e^{3x}cos(3x))/6+(e^{3x}sin(3x))/6+C

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integral of e^{3x}sin(3x)

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

- \frac{e ^{3 x} cos ( 3 x )}{6} + \frac{e ^{3 x} sin ( 3 x )}{6} + C

Solution steps

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

Apply Integration By Parts

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

Take the constant out:

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

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

Apply Integration By Parts

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

Therefore

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

Isolate

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

= - \frac{e ^{3 x} cos ( 3 x )}{6} + \frac{e ^{3 x} sin ( 3 x )}{6}

Add a constant to the solution

= - \frac{e ^{3 x} cos ( 3 x )}{6} + \frac{e ^{3 x} sin ( 3 x )}{6} + C

y^{^{'}} = \frac{x}{y}, y (0) = 3

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\int sin (t^{2}) d t

l a pl a ce t r an s f or m 1 - cos (2 t)

roo t s x sin (y)

What is the integral of e^{3x}sin(3x) ?
The integral of e^{3x}sin(3x) is -(e^{3x}cos(3x))/6+(e^{3x}sin(3x))/6+C