Задача 74061 Найти определенный интеграл методом...
Условие
Решение
du=[m]-\frac{1}{\sqrt{1-(2x)^2}}\cdot 2 \cdot dx[/m]
dv=dx;
v=[m]x[/m]
[m]=( (arccos2x)\cdot (x)|^{\frac{1}{2}}_{-\frac{1}{2}}-∫^{\frac{1}{2}} _{-\frac{1}{2}}x\cdot (-\frac{1}{\sqrt{1-(2x)^2}})\cdot 2dx=[/m]
[m]=(arccos 1)\cdot (\frac{1}{2})-(arccos(-1))\cdot (-\frac{1}{2})+2∫^{\frac{1}{2}} _{-\frac{1}{2}}\frac{(x)}{\sqrt{1-4x^2}}dx=[/m]
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[m]t=1-4x^2[/m]
[m]dt=-8xdx[/m]
[m]xdx=-\frac{1}{8}dt[/m]
[m]∫\frac{(x)}{\sqrt{1-4x^2}}dx=-\frac{1}{8}∫\frac{dt}{\sqrt{t}}=-\frac{1}{8}\cdot 2\sqrt{t}+C=-\frac{1}{4}\sqrt{t}+C=-\frac{1}{4}\sqrt{1-4x^2}+C[/m]
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[m]=0\cdot (\frac{1}{2})-(π)\cdot (-\frac{1}{2})+2\cdot (-\frac{1}{4})(\sqrt{1-4x^2})|{\frac{1}{2}} _{-\frac{1}{2}}=[/m]
[m]=\frac{π}{2}-\frac{1}{2}(\sqrt{1-4\cdot (\frac{1}{2})^2}-\sqrt{1-4\cdot (-\frac{1}{2})^2})=\frac{π}{2}[/m]