Question

Question #4f2f7

Answer 1

#int x^2 sin x dx =sum_(n=0)^infty ((-1)^nx^(2n+4))/((2n+1)!(2n+4))+C#

Explanation:

We know:

`sin x=sum_{n=0}^(infty)((-1)^nx^(2n+1))/((2n+1)!)`

By multiplying by `x^2`,

#x^2 sin x =sum_{n=0}^(infty)((-1)^nx^(2n+3))/((2n+1)!)#

Use Power Rule to integrate term by term,

#int x^2 sin x dx =sum_(n=0)^infty ((-1)^nx^(2n+4))/((2n+1)!(2n+4))+C#