Show, by integrating the Maclaurin series for (x) =11 x 2 , that for |x| sin
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Show, by integrating the Maclaurin series for ƒ(x) =1√1 − x2, that for |x|
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sin ¹ x = x + n=1 1.3.5 (2n-1) x²n+1 2.4.6 (2n) 2n + 1
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