46(е,з) 47(б,г) ..............................

Решите задачу:

$46e)\; \; \int \frac{z^3-8}{2-z}\, dz=\int \frac{(x-2)(z^2+2z+4)}{-(z-2)}\, dz=\int (-z^2-2z-4)\, dz=\\\\=-\frac{z^3}{3}-z^2-4z+C\\\\46z)\; \; \int (u+1)^2(3u-4)du=\int (u^2+2u+1)(3u-4)du=\\\\=\int (3u^3-4u^2+6u^2-8u+3u-4)du=\int (3u^3+2u^2-5u-4)du=\\\\=\frac{3u^4}{4}+\frac{2u^3}{3}-\frac{5u^2}{2}-4u+C\\\\47b)\; \; \int (\frac{cosx}{2}+\frac{1}{cos^2x})dx=\frac{1}{2}\cdot sinx+tgx+C$

$47g)\; \; \int \frac{cos2y}{sin^2y\cdot cos^2y}dy=\int \frac{cos^2y-sin^2y}{sin^2y\cdot cos^2y}dy=\\\\=\int \frac{cos^2y\cdot dy}{sin^2y\cdot cos^2y}-\int \frac{sin^2y\cdot dy}{sin^2y\cdot cos^2y}=\int \frac{dy}{sin^2y}-\int \frac{dy}{cos^2y}=\\\\=-ctgy-tgy+C$