Задача 70998 ...

628a022a6653d96aeebf61ed

24.04.2023 20:36:12

Найти F′(y)

F(y) = ∫ ln(1-y^2x^2)/(x^2sqrt(1-x^2) dx

5f3ea7e3faf909182968ddd9

24.04.2023 23:30:15

★

[m]\frac{d}{dy} F(y)=(∫^{1} _{0}\frac{ln(1-y^2x^2)}{x^2\sqrt{1-x^2}}dx)`_{y}=∫^{1} _{0}(\frac{ln(1-y^2x^2)}{x^2\sqrt{1-x^2}})`_{y}dx=[/m]

[m]=∫^{1} _{0}\frac{1}{x\sqrt{1-x^2}}\cdot \frac{1}{1-y^2x^2}\cdot (1-y^2x^2)`_{y}dx=∫^{1} _{0}\frac{(-2x^2y)}{x\sqrt{1-x^2}\cdot (1-y^2x^2)}dx=-2y∫^{1} _{0}\frac{x}{\sqrt{1-x^2}\cdot (1-y^2x^2)}dx=[/m]

Замена переменной

[m]\sqrt{1-x^2}=t[/m]

[m]x=\sqrt{1-t^2}[/m]

[m]dx=-\frac{2t}{\sqrt{1-t^2}}dt[/m]


[m]=-2y∫^{0} _{1}\frac{\sqrt{1-t^2}}{t\cdot (1-y^2(1-t^2))}(-\frac{2t}{\sqrt{1-t^2}})dt=-2∫^{1} _{0}\frac{d(yt)}{(1-y^2)+(yt)^2}=-2\frac{1}{\sqrt{1-y^2}}arctg\frac{yt}{\sqrt{1-y^2}}|^{1}_{0}=-2\frac{1}{\sqrt{1-y^2}}arctg\frac{y\cdot 1}{\sqrt{1-y^2}}+2\frac{1}{\sqrt{1-y^2}}arctg\frac{y\cdot 0}{\sqrt{1-y^2}}=[/m]

[m]=-2\frac{1}{\sqrt{1-y^2}}arctg\frac{y}{\sqrt{1-y^2}}[/m]