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July 2022 2 13 Report

Помогите решить интеграл: [tex]\int\limits {\frac{dx}{\sqrt{sinxcos^3x} } }[/tex]. Ответ должен быть [tex]2\sqrt{tgx} + C[/tex]

NNNLLL54

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Ответ:

$\displaystyle \int \frac{dx}{\sqrt{sinx\cdot cos^3x}}=\int \frac{\dfrac{dx}{cos^2x}}{\sqrt{\dfrac{sinx\cdot cos^3x}{cos^4x}}}=\int \frac{d(tgx)}{\sqrt{tgx}}=2\sqrt{tgx}+C$

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LymarIvan

Ответ:

Пошаговое объяснение:

$\int {\frac{dx}{\sqrt{\sin x\cos^3x} } } \, dx =\int {\frac{dx}{\sqrt{\frac{\sin x\cos^4x}{\cos x} } } } \, dx =\int {\frac{dx}{\sqrt{\frac{\sin x}{\cos x} *\cos^4x} } } \, =\int {\frac{dx}{\cos^2x\sqrt{\tan x} } } \, dx=$

$\int {\frac{d(\tan x)}{\sqrt{\tan x} } } \, dx =\int {(\tan x)^{-\frac{1}{2} }} \, d(\tan x)=2(\tan x)^\frac{1}{2}+C=2\sqrt{\tan x} +C$

2 votes Thanks 1

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raspishite shagi resheniya

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