Question

How do you integrate `int xsin(4x)` by integration by parts method?

Answer 1

`= - 1/4x cos(4x) + 1/16 sin(4x) + C`

Explanation:

`int xsin(4x) \ dx`

`= int x d/dx ( - 1/4 cos(4x)) \ dx`

which by IBP:
`= - 1/4x cos(4x) + 1/4 int d/dx( x) * cos(4x) \ dx`

`= - 1/4x cos(4x) + 1/4 int cos(4x) \ dx`

`= - 1/4x cos(4x) + 1/4 * 1/4 sin(4x) + C`

`= - 1/4x cos(4x) + 1/16 sin(4x) + C`

Answer 2

`-1/4xcos(4x)+1/16sin(4x)+C`

Explanation:

Before integrating by parts, we can first use substitution to get rid of the `4x` in the sine function.

Let `color(purple)(t=4x`. This implies that `color(green)(dt=4dx` and that `color(blue)(x=t/4`. Thus:

`ttintxsin(4x)dx=1/4intcolor(green)4color(blue)xsin(color(purple)(4x))color(green)(dx)=1/4intt/4sin(t)dt=1/16inttsin(t)dt`

Now is a better time to use integration by parts, which takes the form:

`intudv=uv-intvdu`

So, for `intcolor(red)tcolor(brown)(sin(t)dt`, let:

`color(red)(u=t)" "=>" "du=dt`

`color(brown)(dv=sin(t)dt)" "=>" "intdv=intsin(t)dt" "=>" "v=-cos(t)`

This gives us, following `intudv=uv-intvdu`:

`inttsin(t)dt=-tcos(t)-int(-cos(t))dt`

`=-tcos(t)+intcos(t)dt`

`=-tcos(t)+sin(t)`

Since `t=4x`:

`inttsin(t)dt=-4xcos(4x)+sin(4x)`

And since `intxsin(4x)dx=1/16inttsin(t)dt`, divide the expression by `16`:

`intxsin(4x)dx=-1/4xcos(4x)+1/16sin(4x)+C`