SOLUTION: Evaluate: {{{int( sin(2011(2x-3)pi)*(2x-3)^2011, dx, -1,4)}}}

Algebra ->  Inequalities -> SOLUTION: Evaluate: {{{int( sin(2011(2x-3)pi)*(2x-3)^2011, dx, -1,4)}}}      Log On


   

Question 395544: Evaluate: int%28+sin%282011%282x-3%29pi%29%2A%282x-3%29%5E2011%2C+dx%2C+-1%2C4%29
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
Let u+=+2x+-+3, du+=+2dx. We now have

int%28%28sin%282011u%2Api%29%2Au%5E2011%29%2C+du%29 (I'm writing it at an indefinite integral for now. Don't forget the 1/2 at the end)

Letting y+=+u%5E2011, dz+=+sin+%282011u%2Api%29+du, we obtain dy+=+2011u%5E2010du and z+=+%281%2F%282011%2Api%29%29%2A%28-cos+%282011u%2Api%29%29

Applying integration by parts,



Which is equal to:


Hopefully, by repeatedly using integration by parts, you should be able to get a sum of 2010 or 2011 terms that should be simplifiable in some way. However the easiest way is obviously to use the nint() function or the definite integral function on a graphing calculator.