1) 2) Дано: t=2cS=8 мНайти: S(t)

Решите задачу:

$1.\; \; \int\frac{dx}{sinxcosx}=\int \frac{2dx}{sin2x}=\int\frac{2sin2x\cdot dx}{sin^22x}=-\int \frac{-2sin2x\cdot dx}{1-cos^22x}=-\int \frac{d(cos2x)}{1-cos^22x}=\\\\=[t=cos2x,dt=-2sin2xdx]=-\int\frac{dt}{1-t^2}}=-\frac{1}{2}\cdot ln|\frac{1+t}{1-t}|+C=\\\\=-\frac{1}{2}ln|\frac{1+cos2x}{1-cos2x}|+C=-\frac{1}{2}ln|\frac{2cos^2x}{2sin^2x}|+C=-\frac{1}{2}ln|ctg^2x|+C=\\\\=ln|ctg^2x|^{-\frac{1}{2}}+C=ln|ctgx|^{-1}+C=ln|tgx|+C\\\\1a.\; \; \int \frac{dx}{sinxcosx}=\int\frac{2dx/cos^22x}{tg2x}=$

$=\int\frac{d(tg2x)}{tg2x}=ln|tg2x|+C$

$2.\; \; V=3t^2+6t-4,\; t=2c,\; S=8m\\\\v(t)=S'(t)\quad to\quad S(t)=\int v(t)dt\\\\S(t)=\int (3t^2+6t-4)dt=t^3+3t^2-4t+C\\\\S(2)=8,\\\\S(2)=2^3+3\cdot 4-4\cdot 2+C=12+C,\; \; 12+C=8,\\\\C=8-12=-4\\\\S(t)=t^3+3t^2-4t-4$