int_0^(1/2) arctan(x^2) dx Use a power series to approximate the value of the integral with an error of less than 0.0001.

 From a table of power series, recall that we have:
arctan(x) = sum_(n=0)^oo (-1)^n x^(2n+1)/(2n+1)
To apply this on the given problem, we replace the "x " with "x^2 ".
We get:
arctan(x^2) =sum_(n=0)^oo (-1)^n (x^2)^(2n+1)/(2n+1)
                        =sum_(n=0)^oo (-1)^n x^(2*(2n+1))/(2n+1)
                        =sum_(n=0)^oo (-1)^n x^(4n+2)/(2n+1)
                        = x^2 -x^6/3+x^10/5-x^14/7 +...
The integral becomes:
int_0^(1/2) arctan(x^2)dx = int_0^(1/2) [x^2 -x^6/3+x^10/5- ...]
To determine the indefinite integral, we integrate each term using Power Rule for integration: int x^n dx = x^(n+1)/(n+1) .     
int_0^(1/2) [x^2 -x^6/3+x^10/5-...]
=[x^3/3 -x^7/21+x^11/55-...]_0^(1/2)  
Applying definite integral: F(x)|_a^b = F(b)-F(a) .
F(1/2) or F(0.5)=0.5^3/3 -0.5^7/21+0.5^11/55- ...
                            = 0.0416667 - 0.0003720+0.0000089-...
F(0)=0^3/3 -0^7/21+0^11/55-...
          =0 -0+0 -...     All terms go to zero.
 We stop at the 3rd term since we only need error less than 0.0001 .
int_0^(1/2) arctan(x^2)dx =0.0416667 - 0.0003720+0.0000089
                                = 0.0413036
Thus, int_0^(1/2) arctan(x^2)dx~~0.0413 .