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Answer 1 · 2017-04-10T02:58:41

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

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#