Question

What is `f(x) = int 3sinx-xcosx dx` if `f((7pi)/8) = 0`?

Answer 1

`f(x) = -cosx - xsinx - 3cosx - 2.64`, nearly.

Explanation:

Separate the integrals.

`f(x) = int3sinxdx - intxcosxdx`

Integrate `intxcosxdx` by parts. Let `u = x` and `dv= cosxdx`. Then `du = dx` and `v= sinx`.

`intudv = uv - intvdu`

`intxcosx = xsinx - intsinxdx`

`intxcosx = xsinx - (-cosx) + C`

`intxcosx = xsinx + cosx + C`

Put this together:

`f(x) = int3sinxdx - (xsinx + cosx) + C`

`f(x) = int3sinxdx - xsinx - cosx + C`

`f(x) = C - 3cosx - xsinx - cosx`

You can solve for `C` now. We know that when `x= (7pi)/8`, `y = 0`.

`0 = C - 3cos((7pi)/8) - (7pi)/8sin((7pi)/8) - cos((7pi)/8)`

`C = 3cos((7pi)/8) + (7pi)/8sin((7pi)/8) + cos((7pi)/8)`

This will not be an exact expression. An approximation using a calculator yields `C ~~ -2.64`.

Therefore, `f(x) = -cosx - xsinx - 3cosx - 2.64`.

Hopefully this helps!