Calculus of a Single Variable, Chapter 9, 9.10, Section 9.10, Problem 66

From the table of power series, we have:
cos(x) = sum_(n=0)^oo (-1)^nx^(2n)/((2n))!
= 1-x^2/(2!)+x^4/(4!)-x^6/(6!)+ ...
To apply this on the given integral int_0^1 cos(x^2) dx , we may replace the "x " with "x^2 ".
cos(x^2) = sum_(n=0)^oo(-1)^n (x^2)^(2n)/((2n)!)
=sum_(n=0)^oo (-1)^n x^(4n)/((2n)!)
= 1-x^4/2+x^8/24-x^(12)/720+ x^16/40320 -...
The integral becomes:
int_0^1 cos(x^2) dx =int_0^1 [ 1-x^4/2+x^8/24-x^(12)/720+ x^16/40320-...]dx
To determine the indefinite integral, we integrate each term using Power Rule for integration: int x^ndx =x^(n+1)/(n+1) .
int_0^1 [ 1-x^4/2+x^8/24-x^(12)/720+ x^16/40320-...]dx
=x-x^5/(2*5)+x^9/(24*9)-x^(13)/(720*13)+ x^17/(40320*17)- ...
=x-x^5/10+x^9/216-x^(13)/9360+ x^17/685440- ...
Apply definite integral formula: F(x)|_a^b = F(b) - F(a) .
F(1)=1-1^5/10+1^9/216-1^(13)/9360+ 1^17/685440- ...
=1-1/10+1/216-1/9360+ 1/685440- ...
F(0) =0-0^5/10+0^9/216-0^(13)/9360+ 0^17/685440- ...
= 0-0+0-0+0- ...
All the terms are 0 then F(0)= 0 .
We can stop at 5th term (1/685440~~ 0.0000014589) since we only need an error less than 0.0001 .
Then,
F(1)-F(0)= [1-1/10+1/216-1/9360+ 1/685440-]-[0]
= 0.9045242509
Thus, the approximated integral value is:
int_0^1 cos(x^2) dx ~~0.9045