Question

Va rog ajutati-ma la problema AM 161*

Answer 1

Salut,asa ca idee:

$I=\displaystyle \int \dfrac{1}{tg^2 x-1}dx=\int \dfrac{cos^2 x}{(tg^2 x-1)cos^2 x} dx =\\\text{Dar }cos^2 x=\dfrac{1}{tg^2 x+1} ,deci I=\int \dfrac{1}{\cos^2 x(tg^2 x-1)(tg^2 x+1)}dx\\\\\text{Facem schimbarea de variabila :}u=tg^2 x\Rightarrow du =\dfrac{dx}{cos^2 x}\\\text{Integrala devine }I=\int \dfrac{du}{(u^2-1)(u^2+1)} \\\\\text{Ma gandesc ca te descurci singur de aici (daca nu,nu ezita sa intrebi)}\\\\\\\text{O seara faina!}$