Интеграли

[Гост]

[tex]\int_{}^{ }[/tex] sin2 xdx/1+cos^2 x

[ammornil]

[tex]\int \frac{sin(2x)dx}{1+cos(2x)}=\int\frac{-\frac{1}{2}.(-2).sin(2x)dx}{1+cos(2x)}=-\frac{1}{2}.\int \frac{\[cos(2x)\]'dx}{1+cos(2x)}=-\frac{1}{2}.\int \frac{d\[cos(2x)\]}{1+cos(2x)}=\\
=-\frac{1}{2}.\int \frac{d\[cos(2x)\+1]}{cos(2x)+1}=-\frac{1}{2}.|ln\[cos(2x)+1\]|+C[/tex]

[ammornil]

[tex]I=\int \frac{sin(2x)dx}{1+cos^2 x}=\int \frac{2.sinx.cosx}{1+cos^2 x}dx=2.\int \frac{-[cosx]'.cosx}{1+cos^2 x}dx=-2.\int \frac{cosx}{1+cos^2 x}dcosx \\
\cyr{polagame} \hspace{12} cos x=u, \hspace{12} -1 \le u \le 1 \\
I=-2.\int \frac{u}{1+u^2}du=-2.\int \frac{[\frac{1}{2}.u^2]'du}{1+u^2}=-2.\int \frac{d(\frac{1}{2}u^2)}{1+u^2}=-2.\frac{1}{2}.\int \frac{d(u^2+1)}{u^2+1}=-ln|u^2+1|+C \\
I=-ln|cos^2 x+1|+C[/tex]