1) [tex] \int\limits { \frac{1}{sinxcosx} } \, dx =[/tex]2) Дано: [tex]V=3t^2 +6t -4[/tex]t=2cS=8 мНайти: S(t). Created by o1l7eg17. algebra-ru

tex]t=2cS=8 мНайти: S(t)...

o1l7eg17 @o1l7eg17

July 2022 1 48 Report

1) [tex] \int\limits { \frac{1}{sinxcosx} } \, dx =[/tex]

2) Дано: [tex]V=3t^2 +6t -4[/tex]

t=2c
S=8 м
Найти: S(t)

Answers & Comments

NNNLLL54

Verified answer

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

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

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

2 votes Thanks 1

Answers & Comments

Verified answer

znayushie himiki pomogite reshit pochemu shelochnye metally ne primenyayut dlya vosstan

proshu pomoshi znayushih oharakterizujte elementy metally 2a gruppy psheperiodiche

najti logarifm 28 po osnovaniyu 49 esli logarifm x po osnovaniyu 7m

texnajti s figury

tex]...

reshitsistemyuravnenijspodrobnymresheniemogromnayaprosbanepolzovatsya

perevesti tekst it is necessary to select materials a type a size and a confi

tex]...

vychislit skalyarnoe proizvedenie vektorov aij k i b2i j2k

zachem nuzhno issledovat funkciyu

рекомендуемые вопросы

Helpful Links

Helpful Social