[tex] \int\limits^1_ {-1} \,\frac{arccosx}{1+x^{2}} dx [/tex] va rog daca stiti

Salut,daca notezi integrala cu I si faci schimbarea de variabila x=-t (⇒dx=-dt) obtii:

I=∫ (arccos x)/(1+x^2) dx= ∫ (arccos(-t) )/(1+t²) dt = ∫ (π-arccos t)/(1+t^2) dt =

= ∫ π/(1+t²) dt - I , deci 2I= π ∫ 1/(1+t²) dt ,adica I= π/2 * (arctg 1-arctg (-1) )=

=π/2 *( π/4 -(-π/4))= π/2 * π/2 = π² /4

Daca ai intrebarii,nu ezita sa mi le adresezi.