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July 2022 1 11 Report

Интегрирование методом замены переменно и по частям в неопределенном интеграле

1) [tex] \frac{sin2x}{1+cos^{2} x} [/tex] dx

2) e^-(x^3) dx

3) [tex] \frac{cos \sqrt{x} }{ \sqrt{x} } [/tex] dx

4) [tex] \frac{lnx}{x(1-ln^{2}x) } [/tex] dx

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1) $\int\limits { \frac{sin2x}{1+cos^2 x} } \, dx$

$t = 1+cos^2 x \\ dt = - 2*cosx*sinx*dx = -sin2xdx \\ \\ \int\limits { \frac{sin2x}{1+cos^2 x} } \, dx = -\int\limits { \frac{1}{t} } \, dt = -lnt = -(1+cos^2x)+C$

3) $\int\limits { \frac{cos \sqrt{x} }{ \sqrt{x} } } \, dx$

$t = \sqrt{x} \\ dt = \frac{dx}{2 \sqrt{x} } \\ \\ \int\limits { \frac{cos \sqrt{x} }{ \sqrt{x} } } \, dx = 2\int\limits { cost } \, dt = 2sint = 2sin \sqrt{x}+C$

4) $\int\limits { \frac{lnx}{x(1-ln^2x)} } \, dx$

$t = 1- ln^2x \\ dt = - \frac{2lnx}{x} dx \\ \\ \int\limits { \frac{lnx}{x(1-ln^2x)} } \, dx = - \frac{1}{2} \int\limits { \frac{1}{t} } \, dt = -\frac{1}{2}lnt = -\frac{1}{2}ln(1-ln^2x) +C$

2 votes Thanks 1

Beherith Спасибо

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