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

Question

1274 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-10T01:06:05

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

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!