The taylor series for sin(x) is (will give 10 points)

The Taylor series for sin x about x = 0 is x - (x^3)/3! + (x^5)/5! -.... If f is a function such that f'(x) = sin(x^2) , then the coefficient of x^7 in the Taylor series for f(x) about x = 0 is

The answer is -1/42
I just need to know why... Please show work :)

Substitute x^2 into the Taylor series for sin(x) and you get f'(x) = x^2 - (x^6)/3! + ... Then you can integrate to get f(x) = (x^3)/3 - (x^7)/(7*3!) and 7*3! = 42. So the coefficient is -1/42.

f'(t) = sin(t^2) = t^2 - t^6/6 + t^10/120 +..

Integrate:

f(x) = f(0) + x^3/3 - x^7/42 (+ x^11/1320 +...)