Use series to approximate the definite integral to within the indicated accuracy?

Question

integral (0,.5) cos x^2 dx (SHOW WORK) Give your answer correct to 3 decimal places.

Answer 1

cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! + ...
cos(x^2) = 1 - x^4/2 + x^8/24 - x^12/720 + ...
∫cos(x^2)dx = x - x^5/10 + x^9/216 - x^13/9360 + ...

Since the integral evalutated at 0 is 0, we just need to evaluate this at 0.5

= 0.5 + (0.5)^5/10 + (0.5)^9/216 - ...

= 0.5 - 0.003125 + 0.000009042... - ...
So only the first two terms enter into it for 3 decimal places, since the sum of the 3rd term on is less than 0.000009042.
= 0.5 - 0.003125
= 0.496875 (or 0.496 are the first three decimal digits)
≈ 0.497 rounded to 3 places.

(the exact value is 0.496884033321938...)

Answer 2

cos z= 1-z^2/2!+z^4/4!+++ so cos x^2 = 1-x^4/2!+x^8/4!-
so the integral is
x-x^5/10 +x^9/(9*24)- +-
This is an alternate series and the error in absolute value is less than the first term left out
so as for x= 0 there is no error
(.5)^4n+1/(4n+1)*(2n)! <1/1000
Trial and error
n=2// 0.5^9/9*4! <1/1000 so we have to sum terms n= 0 and n=1
S=0.4969 with three correct decimal figures
If you had to round up it would be 0.497

Answer 3

cosx = a million - (x^2)/2! + (x^4)/4! - (x^6)/6! + . . . replace x via x^3 to get cos(x^3) = a million - (x^6)/2! + (x^12)/4! - (x^18)/6! + . . . Multiply that sequence via the x on the front after which combine it and evaluate as many phrases as needed. i think of that the 1st 3 could do.