Помощ за интеграли :)

[Гост]

[tex]\int_{}^{ }[/tex] x.sin x cos x dx
[tex]\int_{1}^{e}|ln x|dx[/tex] - като интеграла е определен и отгоре е e , а отдолу 1/е

[ammornil]

[tex]\int x.sin x cos x dx =\int x.\frac{1}{2}.2.sinx.cosx dx=\frac{1}{2}.\int x.\frac{1}{2}.2.sin(2x) dx= \frac{1}{4}\int x.\[-cos(2x)\]'dx=\\
=-\frac{1}{4}.\int xd(cos2x)= -\frac{1}{4}.x.cos(2x)+\frac{1}{4}.\int cos(2x)dx=-\frac{1}{4}.x.cos(2x)+\frac{1}{4}.\int cos(2x)d(\frac{1}{2}.2.x)=\\
=-\frac{1}{4}.x.cos(2x)+\frac{1}{8}.\int cos(2x)d2x=-\frac{1}{4}.x.cos(2x)+\frac{1}{8}.sin(2x)+C[/tex]

[ammornil]

[tex]\int_{\frac{1}{e}}^{e}|ln x|dx =? \\
DM: x>0;\\
\int |lnx|dx=x.|lnx|-\int xd(|lnx|)=x.|lnx|-\int x.\frac{1}{|x|}dx=x.|lnx|-x+C=x.(|lnx|-1)+C\\
\int_{\frac{1}{e}}^{e}|ln x|dx =e.(|lne|-1)+C-\frac{1}{e}.(|ln\frac{1}{e}|-1)-C=e.(1-1)-\frac{1}{e}.(|-1|-1)=0[/tex]

[Гост]

[tex]\int_{\frac{1}{e}}^{e}|ln x|dx =e.(|lne|-1)+C-\frac{1}{e}.(|ln\frac{1}{e}|-1)-C=e.(1-1)-\frac{1}{e}.(|-1|-1)=0[/tex]

Tова е революция в интегралното смятане!