Вычислите интеграл [tex]\int\limits^{\frac{\pi }{4}}_0 {\frac{dx}{1+2sin^2x} } \,[/tex]. Created by sobakazabiyaka. algebra-ru

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sobakazabiyaka @sobakazabiyaka

July 2022 1 27 Report

Вычислите интеграл
[tex]\int\limits^{\frac{\pi }{4}}_0 {\frac{dx}{1+2sin^2x} } \,[/tex]

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Ответ:

$\int\limits^{\pi /4}_0\, \dfrac{dx}{1+2sin^2x}=\int\limits^{\pi /4}_0\, \dfrac{dx}{(sin^2x+cos^2x)+2sin^2x}=\int\limits^{\pi /4}_0\, \dfrac{dx}{3sin^2x+cos^2x}=\\\\\\=\Big[\dfrac{:cos^2x}{:cos^2x}\, \Big]=\int\limits^{\pi /4}_0\, \dfrac{\dfrac{dx}{cos^2x}}{3tg^2x+1}=\int\limits^{\pi /4}_0\, \dfrac{d(tgx)}{3tg^2x+1}=\Big[\ t=tgx\ ,\ t_1=0\ ,\ t_2=1\Big]=$

$=\int\limits^{1}_0\, \dfrac{dt}{3t^2+1}=\dfrac{1}{\sqrt3}\cdot \int\limits^{1}_0\, \dfrac{\sqrt3\, dt}{(\sqrt3t)^2+1}=\Big[\ u=\sqrt3t\ ,\ u_1=0\ ,\ u_2=\sqrt3\ \Big]=\\\\\\=\dfrac{1}{\sqrt3}\cdot \int\limits^{\sqrt3}_0\, \dfrac{du}{u^2+1}=\dfrac{1}{\sqrt3}\cdot arctg\, u\Big|_0^{\sqrt3}=\dfrac{1}{\sqrt3}\cdot (arctg\sqrt3-arctg0)=\\\\\\=\dfrac{1}{\sqrt3}\cdot \dfrac{\pi}{3}=\dfrac{\pi}{3\sqrt3}$

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