d)
[tex]\int \frac{sin^{3}x-8}{1-cos^{2}x} \, dx[/tex] = [tex]\int \frac{sin^{3}x-8}{sin^{2}x} \,dx[/tex] = [tex]\int sinx\,dx - 8\int\frac{1}{sin^{2}x}\,dx[/tex] = -cosx-ctgx+C
e)
Folosim schimbare de variabila : 2x = t => dx = 1/2 dt
[tex]\frac{1}{2}\int\frac{3cost-1}{sin^{2}t}\,dt[/tex] = [tex]\frac{3}{2}\int\frac{cost}{sin^{2}t}\,dt - \frac{1}{2}\int\frac{1}{sin^{2}t}\,dt[/tex]
Pentru a rezolva prima integrala folosim schimbarea de variabila : sin t = u => dt = 1/cos t du
Obtinem [tex]\frac{3}{2}\int\frac{1}{u^{2}}\,du - \frac{1}{2}\int\frac{1}{sin^{2}t}\,dt[/tex]
Rezultatul final : [tex]-\frac{3}{2sin2x}-\frac{ctg2x}{2}+C[/tex]
