There are a lot of pies here (My tenth integral problem)

by tabular integration,we can notice a pattern

$\displaystyle\int { x^{ n }\sin { (x) } } =cos(x)\displaystyle\sum _{ k=0 }^{ \left\lfloor \frac { n }{ 2 } \right\rfloor }{ (-1)^{ k+1 }\frac { n!x^{ n-2k } }{ (n-2r)! } } +\sin (x)\displaystyle\sum _{ k=0 }^{ \left\lfloor \frac { n-1 }{ 2 } \right\rfloor }{ (-1)^{ k } } \frac { n!{ x }^{ n-2k-1 } }{ (n-2k-1)! } \\ \\$

evaluating the integral at x= $\frac{\pi}{2}$ we get the highest power is from the second sum which is 2016-1= $\boxed{2015}$ then evaluating it at x=0 we get 0 since x is in both sums it zeros every term in the sum so the answer is

$\boxed{2015}$

nice problem!

Bonus :

we can derive the general formula for cosine using the same method

it is

$\displaystyle\int { x^{ n }\cos { (x) } dx } =\sin { (x) } \displaystyle\sum _{ i=0 }^{ \left\lfloor \frac { n }{ 2 } \right\rfloor }{ (-1)^{ i } } \frac { n!x^{ m-2i } }{ (n-2i)! } +\cos (x)\displaystyle\sum _{ i=0 }^{ \left\lfloor \frac { n-1 }{ 2 } \right\rfloor }{ (-1)^{ i } } \frac { n!{ x }^{ n-1-2i } }{ (n-2i-1)! }$

so in this case the highest power is 2016,or generally, $n$

this works because of integration by parts.I think that there may be a combinatoric explanation since i see factorials and even more so ,i think that there may be a geometrical interpretation.If anyone wants to do another approach,that would be awesome because i am eager to see it :)