Răspuns:
Iti las un alt mode de a rezolva:
[tex]\displaystyle\int\dfrac{1}{\cos x}dx=\int\dfrac{\cos x}{\cos^2 x}dx=\int\dfrac{\cos x}{1-\sin^2x}dx=\int\dfrac{1}{1-t^2}dt=\\\sin x=t\Rightarrow \cos xdx=dt\\=-\dfrac{1}{2}\ln\left|\dfrac{t-1}{t+1}\right|+C=-\dfrac{1}{2}\ln\left|\dfrac{\sin x-1}{\sin x+1}\right|+C[/tex]
Explicație pas cu pas:
