Question

Use integration by parts to find `int xsin(pix) dx`?

Answer 1

`int \ xsin(pix) \ dx = sin(pix)/pi^2 - (xcos(pix))/pi + C`

Explanation:

We could use integration by parts or try a bit of guess work by trying some suitable function to differentiate and seeing if we can find the solution.

Consider how the product rule works:

`d/dx (uv) = u (dv)/dx + (du)/dxv`

So if we tried `y=xcos(ax)` we will end up with something close to what we need:

`y=xcos(ax)`

Differentiate wrt `x`:

`dy/dx=x(d/dx cos(ax)) + d/dx(x)(cos(ax))`
`dy/dx=-axsin(ax) + cos(ax)`

So we have:

`d/dx (xcos(ax)) = cos(ax)-axsin(ax)`

In other words (by FTOC):

`int \ cos(ax)-axsin(ax) \ dx = xcos(ax) + c`

`:. int \ cos(ax) \ dx - int \ axsin(ax) \ dx = xcos(ax) + c`

`:. 1/asin(ax) - int \ axsin(ax) \ dx = xcos(ax) + c`

`:. int \ axsin(ax) \ dx = 1/asin(ax) - xcos(ax) - c`

`:. int \ xsin(ax) \ dx = sin(ax)/a^2 - (xcos(ax))/a - c/a`

`:. int \ xsin(ax) \ dx = sin(ax)/a^2 - (xcos(ax))/a + C`

Hence:

`int \ xsin(pix) \ dx = sin(pix)/pi^2 - (xcos(pix))/pi + C`