The discussion focuses on solving the integral of arctg(x) divided by sqrt(1-x^2). Participants explore substitution methods, particularly using u = arctg(x) and du = 1/(1+x^2), but express skepticism about the existence of an antiderivative due to Wolfram Alpha's inability to find one. A suggestion is made to utilize the MacLaurin series expansion for arctg(x), which yields a series representation that can be integrated term by term. However, there is a request for a solution using traditional methods, as some students have not yet studied series.
PREREQUISITESStudents studying calculus, particularly those learning integration techniques, and educators seeking to provide traditional methods for solving integrals involving trigonometric functions.
leprofece said:integral of arctg(x)/sqrt(1-x^2)
maybe u = arctg x du = 1 /1+x^2
but x = tg u
maybe this is the way isn't it?
chisigma said:A possible approach is to use the MacLaurin expansion...
$\displaystyle \frac{\tan^{-1} x}{\sqrt{1-x^{2}}} = x + \frac{1}{6}\ x^{3} + \frac{49}{120}\ x^{5} + \frac{81}{560}\ x^{7} + \mathcal{O}\ (x^{9})\ (1)$
... and to integrate term by term...
Kind regards $\chi$ $\sigma$
leprofece said:Thank for your answer but you must solve it by normal or traditional methods students have not studied series yet
Could anybody do that way??