1 Answer | Add Yours
We know that `(d(arctgx))/(dx)=1/(1+x^2)`
Hence, the best pattern for Taylor's series for `arctg x` , around `x = 0` , can be found from the series for `1/(1+x^2)` .
So, `(d(arctgx))/(dx)` =`1/(1+x^2)` =`1-x^2+x^4-x^6+x^8` `-` .......for` -1ltxlt1`
Integrating term by term gives:
`arctgx` =`int 1/(1+x^2)`
`=C+x-x^3/3+x^5/5-x^7/7+x^9/9` ..............for `-1ltxlt1` , where `C ` is the constant of integration.
Since, `arc tg0=0` we have `C=0` , so,
`arctgx=x-x^3/3+x^5/5-x^7/7+x^9/9` ......for `-1ltxlt1` .
We’ve answered 319,852 questions. We can answer yours, too.Ask a question