Question

1) $intlimits { frac{1}{sinxcosx} } , dx =$

2) Дано: $V=3t^2 +6t -4$

t=2c
S=8 м
Найти: S(t)

Answer 1

$1.; ; intfrac{dx}{sinxcosx}=int frac{2dx}{sin2x}=intfrac{2sin2xcdot dx}{sin^22x}=-int frac{-2sin2xcdot dx}{1-cos^22x}=-int frac{d(cos2x)}{1-cos^22x}=\\=[t=cos2x,dt=-2sin2xdx]=-intfrac{dt}{1-t^2}}=-frac{1}{2}cdot ln|frac{1+t}{1-t}|+C=\\=-frac{1}{2}ln|frac{1+cos2x}{1-cos2x}|+C=-frac{1}{2}ln|frac{2cos^2x}{2sin^2x}|+C=-frac{1}{2}ln|ctg^2x|+C=\\=ln|ctg^2x|^{-frac{1}{2}}+C=ln|ctgx|^{-1}+C=ln|tgx|+C\\1a.; ; int frac{dx}{sinxcosx}=intfrac{2dx/cos^22x}{tg2x}=$

$=intfrac{d(tg2x)}{tg2x}=ln|tg2x|+C$

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