Question

Va rog mult AJUTOR! Poza mai jos cu niste integrale. Doar alea cu galben

Answer 1

d)

$\int \frac{sin^{3}x-8}{1-cos^{2}x} \, dx$ = $\int \frac{sin^{3}x-8}{sin^{2}x} \,dx$ = $\int sinx\,dx - 8\int\frac{1}{sin^{2}x}\,dx$ = -cosx-ctgx+C

e)

Folosim schimbare de variabila : 2x = t => dx = 1/2 dt

$\frac{1}{2}\int\frac{3cost-1}{sin^{2}t}\,dt$ = $\frac{3}{2}\int\frac{cost}{sin^{2}t}\,dt - \frac{1}{2}\int\frac{1}{sin^{2}t}\,dt$

Pentru a rezolva prima integrala folosim schimbarea de variabila : sin t = u => dt = 1/cos t du

Obtinem $\frac{3}{2}\int\frac{1}{u^{2}}\,du - \frac{1}{2}\int\frac{1}{sin^{2}t}\,dt$

Rezultatul final : $-\frac{3}{2sin2x}-\frac{ctg2x}{2}+C$