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

1 Answer
Feb 10, 2017

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!