Use integration by parts to evaluate the integral

Z

e

$3$x

sin$(3$x$)$dx

Notice that this is an integration by parts problem in which you are

required to do integration by parts twice and then do some algebra

to get the answer.

To get started recall the differential form of the integration by parts

formula used in the book

Z

u dv $=$ uv$$

Z

v du

In order to get started you need to choose values for u and dv$.$ The

values you choose will determine how to proceed.

u $=$ e

$3$x or u $=$ sin$(3$x$)$

If you choose u $=$ e

$3$x

then you will have

du $=$ dx$,$

dv $=$ dx$,$

v $=,$

Inserting these expressions in the integration by parts formula and

simplifying you obtain

R

e

$3$x

sin$(3$x$)$dx $=+$

R

e

$3$x

cos$(3$x$)$dx

Next we need to do integration by parts again and, ignoring all the

other terms, the integral you need to evaluate is

Z

e

$3$x

cos$(3$x$)$dx

Because of your choice for u above you now are forced to use the

same type of choice used above and you would set

u $=,$

du $=$ dx$,$

dv $=$ dx$,$

v $=,$

Collecting these results you now have

R

e

$3$x

sin$(3$x$)$dx $=+$

R

e

$3$x

sin$(3$x$)$dx

Finally move the last term on the right to the other side, collect

terms and solve for the desired integral to obtain

R

e

$3$x

sin$(3$x$)$dx $=$