\[{\text{Докажите}}{\text{, что }}\int\limits_0^{2\pi } {\frac{{x\sin x + \cos x}}{{{{\cos }^2}x + {x^2}}}dx} = \operatorname{arctg} 2\pi .\]

\[{\text{Докажите}}{\text{, что }}\int\limits_0^{\frac{\pi }{2}} {\frac{{dx}}{{{{\sin }^{10}}x + {{\cos }^{10}}x}} = \sqrt 5 \pi .} \]

\[\int\limits_0^{\frac{\pi }{2}} {\left( {\prod\limits_{k = 1}^8 {\sin \left( {kx} \right)} } \right)dx} = \frac{\pi }{{256}}.\]

\[\int\limits_0^{\frac{\pi }{2}} {\sin \left( {2020x} \right)\sin \left( {2021x} \right)dx} \]

\[\int\limits_0^1 {\ln x\ln \left( {1 - x} \right)dx} = 2 - \frac{{{\pi ^2}}}{6}.\]

\[\int\limits_0^1 {\frac{{\ln x\ln \left( {1 - x} \right)}}{x}dx} = \zeta \left( 3 \right).\]

\[\begin{array}{l} \int\limits_0^n {{{\left( {1 + x} \right)}^x}dx} \hfill \\ n \in \mathbb{N} \hfill \\ \end{array}\]

\[\int\limits_0^1 {\int\limits_0^1 {\int\limits_0^1 {\frac{{\ln \left( {xyz} \right) \cdot \ln \left( {1 - xyz} \right)}}{{xyz}}} } } dzdydx = 3\zeta \left( 5 \right)\]

\[\int\limits_0^1 {{x^{ - x}}dx} = \sum\limits_{k = 1}^{ + \infty } {{k^{ - k}}} \]

\[\begin{array}{l} \int\limits_0^1 {{x^a}{{\ln }^{2k}}xdx} \hfill \\ k \in \mathbb{N},{\text{ }}a \geqslant 0 \hfill \\ \end{array}\]

\[\int\limits_0^\pi {\ln \left( {1 + \frac{1}{{\sin x}}} \right)dx} \]

\[\int\limits_0^{\frac{\pi }{2}} {\frac{x}{{\sin x}}dx} \]

\[\int\limits_0^{\frac{\pi }{2}} {\frac{{{x^2}}}{{{{\sin }^2}x}}dx} \]

\[\int\limits_0^{\frac{\pi }{4}} {{{\ln }^4}\left( {\operatorname{tg} x} \right)dx} \]

\[\int\limits_0^1 {\frac{{\arctan \left( {{x^k}} \right)}}{x}dx} \]

\[\int\limits_0^1 {\frac{{{{\ln }^2}x}}{{1 + {x^2}}}dx} \]

\[\int\limits_0^{\frac{\pi }{4}} {\ln \left( {\operatorname{tg} x} \right)dx} \]

\[\begin{array}{l} {\text{Докажите}}{\text{, что число }}\int\limits_0^1 {\sqrt[n]{{1 - \sqrt[n]{{1 - \sqrt[n]{{1 - x}}}}}}} dx{\text{ - рациональное}}{\text{.}} \hfill \\ (n = 2,3,4,...) \hfill \\ \end{array}\]

\[{\text{Найдите значение интеграла: }}\int\limits_0^1 {\sqrt[n]{{1 - \sqrt[n]{{1 - x}}}}dx} ,{\text{ }}n = 2,3,4,...\]

\[\begin{array}{l} {\text{Вычислите:}} \hfill \\ \mathop {\lim }\limits_{a \to 0 + } \left( {{a^{\frac{5}{3}}}\int\limits_a^{\frac{\pi }{2} - a} {\frac{{dx}}{{{{\left( {1 - \sin x} \right)}^{\frac{4}{3}}}}}} } \right) \hfill \\ \end{array}\]

$%\int\limits_0^{\frac{\pi }{2}} {\frac{1}{{1 + {{\operatorname{tg} }^s}x}}dx} ,{\text{ }}s \in \mathbb{R}$%

\[\begin{array}{l} \int\limits_0^{\frac{\pi }{2}} {{{\sin }^{2k + 1}}xdx} = \frac{{\left( {2k} \right)!!}}{{\left( {2k + 1} \right)!!}}, \hfill \\ \int\limits_0^{\frac{\pi }{2}} {{{\sin }^{2k}}xdx} = \frac{{\left( {2k - 1} \right)!!}}{{\left( {2k} \right)!!}} \cdot \frac{\pi }{2}, \hfill \\ k \in \mathbb{N} \hfill \\ \end{array} \]

\[\int\limits_0^{\frac{\pi }{2}} {{{\operatorname{tg} }^s}x} dx,{\text{ }}0 < s < 1.\]

\[\int\limits_0^{\frac{\pi }{2}} {\int\limits_0^{\frac{\pi }{2}} {\frac{{\sin x\sin y}}{{{{\sin }^2}x + {{\sin }^2}y}}dxdy} } = Catalan.\]

\[\int\limits_0^1 {\Gamma \left( {2 - x} \right)\Gamma \left( {2 + x} \right)dx} = \frac{{21\zeta \left( 3 \right)}}{{2{\pi ^2}}}.\]

\[\int\limits_0^{\frac{1}{2}} {\frac{{\Gamma \left( {1 - x} \right)\Gamma \left( {1 + x} \right)}}{{\Gamma \left( {1 - \frac{x}{2}} \right)\Gamma \left( {1 + \frac{x}{2}} \right)}}dx} \]

\[{\text{Let }}a,b,c \in \mathbb{R},{\text{ }}0 < a < b.{\text{ Show that}}\] $$\int\limits_0^1 {{{\left( {{x^a} - {x^b}} \right)}^c}dx} = \frac{{\Gamma \left( {\frac{{ac + 1}}{{b - a}}} \right) \cdot \Gamma \left( {c + 1} \right)}}{{\Gamma \left( {\frac{{bc + 1}}{{b - a}}} \right) \cdot \left( {bc + 1} \right)}}.$$

\[\begin{array}{l} \int\limits_0^1 {\frac{{\ln {{\left( x \right)}^m}}}{{1 - {x^s}}}dx} = - \frac{1}{{{s^{m + 1}}}}\Psi \left( {m,\frac{1}{s}} \right), \hfill \\ s > 0,{\text{ }}m \in \mathbb{N} \hfill \\ \end{array} \]

\[\int\limits_0^1 {\ln \Gamma \left( x \right)dx} \]

$$\int\limits_0^{\frac{{2\pi }}{3}} {{{\ln }^2}\left( {2\cos \frac{x}{2}} \right)dx} $$