Помогите с математикой :(. Created by Asagami. matematika-ru

11)t=3+2sinx⇒dt=2cosxdx⇒cosxdx=dt/22)3)2y`(π/4)=-4/sinπ/2=-4/1=-43y`=4x³-4=4(x³-1)=0x³=1x=1 _ +------------------(1)---------------------убыв x∈(-∞;1)возр x∈(1;∞)

Помогите с математикой :(...

Asagami @Asagami

August 2022 1 9 Report

Помогите с математикой :(

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sedinalana

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1
1)t=3+2sinx⇒dt=2cosxdx⇒cosxdx=dt/2
$\int\limits {cosx/(3+2sinx)} \, dx =1/2 \int\limits {1/t} \, dt =1/2*lnt=1/2*ln(3+2sinx)+$ $C$
2) $\int\limits1/(3x+1)^2 \, dx =-1/3(3x+1)+C$
3) $\int\limits {4/cos^2x} \, dx =4tgx| \pi /4-(- \pi /4)=4tg \pi /4-4tg(- \pi /4)=4+4=8$
2 $y`=1/ctg^2x*2ctgx*(-1/sib^2x)=-2/(ctgx*sin^2x)=-2/(cosxsinx$ $)=-4/sin2x$
y`(π/4)=-4/sinπ/2=-4/1=-4
3
y`=4x³-4=4(x³-1)=0
x³=1
x=1
_ +
------------------(1)---------------------
убыв x∈(-∞;1)
возр x∈(1;∞)

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