# `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)`

...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

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` .

Approved by eNotes Editorial Team