Question #4f2f7

1 Answer
Wataru
Apr 10, 2017

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

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