[tex]I= \int\limits{cosxln(1+cosx)} \, dx = ?[/tex]
Vom rezolva aceasta integrala prin parti , unde f derivat va fi cosx si g(x) va fi ln(1+cosx) .
[tex]f'(x)=cosx ==> f(x) = sinx \\ g(x) = ln(1+cosx) ==> g'(x) = - \frac{sinx}{1+cosx} \\ \\ I = sinxln(1+cosx) + \int\limits { \frac{ sin^{2}x }{1+cosx} } \, dx = sinxln(1+cosx) + \int\limits { \frac{1- cos^{2}x }{1+cosx} } \, dx [/tex]
Am folosit formula : [tex] sin^{2}x + cos^{2}x = 1 [/tex]
[tex]I = sinxln(1+cosx) + \int\limits{ \frac{ (1-cosx)(1+cosx)}{1+cosx} } \, dx [/tex]
[tex]\\ I= sinxln(1+cosx) + \int\limits {(1-cosx)} \, dx \\ \\ I= sinxln(1+cosx) +x-sinx+C[/tex]
