Вычислить определенный интеграл:

Решите задачу:

$1)\quad \int\limits _{\frac{1}{\sqrt{e}}}^{e}\frac{dx}{x}=ln|x|\limits |^{e}_{\frac{1}{\sqrt{e}}}=lne-ln\frac{1}{\sqrt{e}}=1-(ln1-ln\sqrt{e})=\\\\=1-(0-\frac{1}{2})=\frac{1}{2}\\\\2)\quad \int\limits^0_{-\frac{\pi}{2}} {sin(\frac{\pi}{4}-3x)} \, dx =-\frac{1}{3}\cdot (-cos(\frac{\pi}{4}-3x))|_{-\frac{\pi}{2}}^0=\\\\=\frac{1}{3}\left (cos\frac{\pi}{4}-cos(\frac{\pi}{4}+\frac{3\pi}{2})\right )=\frac{1}{3}\left (\frac{\sqrt2}{2}-cos\frac{7\pi}{4}\right )=$

$=\frac{1}{3}\left (\frac{\sqrt2}{2}-\frac{\sqrt2}{2}\right )=0$

$3)\quad \int \limits _0^1\, arctgx\, dx=[\, u=arctgx,\; du=\frac{dx}{1+x^2}\, ,\, dv=dx\; ,\; v=x]=\\\\=x\cdot arctgx|_0^{\frac{\pi}{4}}-\int \limits _{0}^{\frac{\pi}{4}}\, \frac{x\, dx}{1+x^2} =\frac{\pi}{4}\cdot arctg\frac{\pi}{4}-\frac{1}{2}\cdot \int \limits _0^{\frac{\pi}{4}}\frac{2x\, dx}{1+x^2}=\\\\=\frac{\pi}{4}\cdot arctg\frac{\pi}{4}-\frac{1}{2}\cdot ln|1+x^2||_0^{\frac{\pi}{4}}=\\\\=\frac{\pi}{4}\cdot arctg\frac{\pi}{4}-\frac{1}{2}(ln(1+\frac{\pi ^2}{16})-ln1)=$

$=\frac{\pi}{4}\cdot arctg\frac{\pi}{4}-\frac{1}{2}ln(1+\frac{\pi ^2}{16})$