MHB Integral by Parts: Solving 2 Integrals Involving Arctg(x) & Sqrt(1-x^2)

[leprofece]

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?

 

[Prove It]

Wolfram couldn't find an antiderivative so I doubt there is one...

 

[chisigma]

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?

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]

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 $\chi$ $\sigma$

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??

 

[DreamWeaver]

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??

Hey! :D

As with your other recent thread - for $$\int \sqrt{x}e^x\, dx$$ - it sounds like your teacher is asking you to do stuff without him or her teaching you the things necessary to solve the problem... (Headbang)