Пошаговое объяснение:

3.

[tex]\displaystyle\\\int(\frac{2}{sin^2(4x)}+\frac{6}{2x-3})dx=\int\frac{2}{sin^2(4x)} dx+\int\frac{6}{2x-3}dx.\\\\ a)\ \int\frac{2}{sin^2(4x)}dx=\left [ {{u=4x \ \ \ \ x=\frac{u}{4}} \atop {dx=\frac{du}{4} }} \right ] =\int\frac{2}{sin^2u}\frac{du}{4} =\frac{1}{2} \int\frac{du}{sin^2u} =\\\\=-\frac{ctg(u)}{2} +C=[ \ u=4x ]=-\frac{ctg(4x)}{2} +C)\ .\\\\[/tex]

[tex]\displaystyle\\b)\ \int\frac{6}{2x-3}dx=\left [ {{2x-3=u\ \ \ \ x=\frac{u+3}{2} } \atop {dx=\frac{du}{2} }} \right]=6*\int\frac{1}{u}*\frac{du}{2} =3*\int\frac{du}{u} =\\\\ =3*ln|u|+C=[u=2x-3]=3*ln|2x-3|+C.\ \ \ \ \Rightarrow\\\\\int(\frac{2}{sin^2(4x)}+\frac{6}{2x-3})dx=3*ln|2x-3|-\frac{ctg(4x)}{2}+C .[/tex]

4.

[tex]\displaystyle\\\int\limits^\frac{\pi}{16} _0 {\frac{1}{cos^2(8x)} } \, dx=\infty.[/tex]

[tex]\displaystyle\\\int\frac{1}{cos^2(8x)} dx=\left [ {{u=8x\ \ \ \ x=\frac{u}{8} } \atop {dx=\frac{du}{8} }} \right]=\int\frac{1}{cos^2u} *\frac{du}{8}=\\\\=\frac{tg(u)}{8}=[u-8x]=\frac{tg(8x)}{8}+C.\\\\ \frac{tg(8x)}{8} \ |^\frac{\pi}{16}_0 =\frac{1}{8}*(tg(\frac{8\pi }{16})-tg(8*0))=\frac{1}{8} *(tg\frac{\pi }{2} -tg0)=\frac{1}{8}*(\infty-0)=\infty.[/tex]

Ответ: интеграл расходится.