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Объяснение:

[tex]1)\\\int\limits {\frac{32x}{(x^2+1)^5} } \, dx =\boxed {\left {{u=x^2+1} \atop {du=2xdx\ |:2\ \ \ \ \frac{du}{2}=xdx }} \right. }=\int\limits {\frac{32}{2*u^5} } \, du =\int\limits {16u^{-5}du} =\\ =16*\int u^{-5}du=16*\frac{u^{-4}}{-4} =-\frac{4}{u^4} =-\frac{4}{(x^2+1)^4}.\\[/tex]

[tex]{-\frac{4}{(x^2+1)^4} } \ |_1^{\sqrt3 }=-(\frac{4}{((\sqrt3) ^2+1)^4}-\frac{4}{(1 ^2+1)^4})=-(\frac{4}{(3+1)^4}-\frac{4}{(1+1)^4}) =-(\frac{4}{4^4} -\frac{4}{2^4})=\\ =-(\frac{1}{4^3}-\frac{4}{16})=-(\frac{1}{64}-\frac{1}{4})=-\frac{1-16}{64} =-\frac{-15}{64}=\frac{15}{64}.[/tex]

[tex]2)\\\int\limits {cos\frac{x}{2} } \, dx =\boxed {\frac{x}{2}=u\ \ \ \ \frac{dx}{2}=du\ |*2\ \ \ \ dx=2du }=\int 2*cosudu=\\=2sinu=2sin\frac{x}{2} .\\2sin\frac{x}{2} \ |_0^\frac{2\pi}{3} =2*(sin(\frac{\frac{2pi}{3} }{2} )-sin\frac{0}{2})=2*(sin\frac{\pi}{3}-sin0)=2*\frac{\sqrt{3} }{2} -2*0=\sqrt{3} .[/tex]