`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` .
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