maclaurin series calculus 2?

the maclaurin series for sin x is series from n=0 to infiinity is ( (-1)^n*(x^(2n+1)/(2n+1)!).

Find the maclaurin series for sin 3x.

find the maclaurin series for the integral from 0 to x of sin t^2 dt.

uuse the maclaurin series for sinx to approximate sin 1/2 with four decimal precision

Comments

  • sin 3x = sum from n=0 to infiinity is ( (-1)^n*((3x)^(2n+1)/(2n+1)!)

    integral from 0 to x of sin t^2 dt

    = integral from 0 to x of ( (-1)^n*((t^2)^(2n+1)/(2n+1)!) dt

    = [(-1)^n/(4n+3)]t^(4n+3)/(2n+1)! from 0 to x

    = [(-1)^n/(4n+3)]x^(4n+3)/(2n+1)!

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