View Single Post
Anamitra
Anamitra is offline
#58
Jul10-11, 08:56 PM
P: 621
An Alternative Treatment

[tex]\frac{1}{{1}{+}{x}^{2}}{=}\frac{1}{2i}{[}\frac{1}{x-i}{-}\frac{1}{x+i}{]}[/tex]

[tex]{=}{-}\frac{1}{2i}{[}\frac{1}{i-x}{+}\frac{1}{i+x}{]}[/tex]

[tex]{=}{-}\frac{1}{2i}{i}^{-1}{[}{{(}{1}{-}{x}{/}{i}{)}}^{-1}{+}
{{(}{1}{+}{x}{/}{i}{)}}^{-1}{]}[/tex]

Applying the binomial expansion and after cancellations we have:

Integrand=[tex]{[}{1}{-}{x}^{2}{+}{x}^{4}{-}{x}^{6}............{]}[/tex]

On integration we have,
Integral=[tex]{[}{x}{-}{{x}^{3}}{/}{3}{+}{{x}^{5}}{/}{5}{-}{{x}^{7}}{/}{7} ..........{]}[/tex]
= arctan{x}

[link for the expansion of arc tan(x): http://en.wikipedia.org/wiki/Taylor_series ]
Attached Images
File Type: bmp maths_post.bmp (235.1 KB, 7 views)