Интеграли

[alexandrina__]

[tex]\int_{}^{ } cos^3xdx[/tex] =?

[tex]\int_{}^{ } x^2 \sqrt{} 1-x^2dx[/tex] ( 1- [tex]x^2[/tex] е под корен)

[tex]\int_{}^{ }[/tex] xln (1+x) dx

[tex]\int_{}^{ }[/tex] [tex]x^3[/tex] [tex]e^2^x[/tex] dx

[tex]\int_{}^{ }[/tex] [tex]x^2[/tex] arctgxdx

[Mark]

[tex]\int_{}^{ } {x^3e^{2x}dx}=\frac{1}{ 2} x^3e^{2x}-\frac{3}{2 }\int_{}^{ } {e^{2x}x^2dx}....[/tex]
[tex]\int_{}^{ } {xln(x+1)dx}=\frac{1}{2 } x^2ln(x+1)-\frac{1}{ 2} \int_{}^{ } {\frac{x^2}{x+1 }dx}[/tex] [tex]x+1=t[/tex] [tex]dx=dt[/tex]...
[tex]\int_{}^{ } {x^2arctan(x)dx}=\frac{x^3}{ 3} arctan(x)-\frac{1}{ 3} \int_{}^{ } {\frac{x^3}{ 1+x^2}dx}[/tex]
[tex]\int_{}^{ } {\frac{x^3}{ 1+x^2} dx}=\int_{}^{ } {x-\frac{x}{1+x^2 }dx}[/tex] [tex]1+x^2=t[/tex] [tex]dx=\frac{dt}{2\sqrt{t-1} }[/tex]
[tex]\int_{}^{ } {\frac{\sqrt{t-1}dt }{2t\sqrt{t-1} }=\frac{1}{ 2} \int_{}^{ } {\frac{1}{t } dt}[/tex] [tex]=>[/tex]
[tex]\int_{}^{ } {x^2arctan(x)dx}=\frac{x^3}{ 3} arctan(x)-\frac{1}{ 3}\int_{}^{ } {xdx} -\frac{1}{6 }\int_{}^{ } {\frac{1}{ t}dt}[/tex]
[tex]\int_{}^{ } {x^2\sqrt{1-x^2}dx }[/tex] [tex]x\in [-1;1] x=sint[/tex]
[tex]\int_{}^{ } {cos^3xdx}=\frac{1}{ 4} \int_{}^{ } {(cos3x+3cosx)dx}[/tex]