Задача 58691 пожалуйста помогите ...
Условие
Решение
[m] ∫^{1}_{\frac{1}{2}} \frac{x}{(1-x^2)^{\frac{1}{2}}}dx[/m]
Табличный интеграл
[m] ∫ \frac{du}{\sqrt{u}}=2\sqrt{u}+C[/m]
[m]u=1-x^2[/m]
[m]du=(1-x^2)`dx=-2xdx[/m]
[m] ∫^{1}_{\frac{1}{2}} \frac{x}{(1-x^2)^{\frac{1}{2}}}dx=-\frac{1}{2}∫^{1}_{\frac{1}{2}} \frac{(-2xdx)}{\sqrt{1-x^2}}=-\frac{1}{2}∫^{1}_{\frac{1}{2}} \frac{d(1-x^2)}{\sqrt{1-x^2}}=-\frac{1}{2}\cdot 2\sqrt{1-x^2}|^{1}_{\frac{1}{2}}=[/m]
[m]=-\frac{1}{2}\cdot \sqrt{1-1^2}+\frac{1}{2}\cdot \sqrt{1-(\frac{1}{2})^2}=...[/m] считайте
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Тригонометрические подстановки.
[m]x=3 tgt[/m]
[m]dx=3 (tgt)`dt[/m]
[m]dx=\frac{3}{cos^2t}dt[/m]
[m]9+x^2=9+9tg^2t=9\cdot (1+tg^2t)=\frac{9}{cos^2t}[/m]
[m]\sqrt{9+x^2}=\frac{3}{cost}[/m]
Тогда
[m]∫ \frac{dx}{x^2\cdot \sqrt{9+x^2}}= ∫ \frac{\frac{3}{cos^2t}dt}{(9tg^2t\cdot \frac{3}{cost}}dx= 3∫\frac{cost}{sin^2t}dt=-3\frac{1}{sint}+C [/m]
так как
[m]x=3 tgt[/m] ⇒ [m]tgt=\frac{x}{3}[/m]
[m]1+tg^2t=1+(\frac{x}{3})^2[/m]
[m]\frac{1}{cos^2t}=1+\frac{x^2}{9}[/m]
[m]\frac{1}{cos^2t}=\frac{9+x^2}{9}[/m] ⇒ [m]cos^2t=\frac{9}{9+x^2}[/m]
[m]sin^2t=1-cos^2t[/m]
[m]sin^2t=1-\frac{9}{9+x^2}[/m] ⇒ [m]sin^2t=\frac{x^2}{9+x^2}[/m]
[m]sint=\sqrt{\frac{x^2}{9+x^2}}[/m]
[m]\frac{1}{sint}=\frac{\sqrt{9+x^2}}{x}[/m]
Окончательно:
[m]∫ \frac{dx}{x^2\cdot \sqrt{9+x^2}}= -3\frac{1}{sint}+C=-3\frac{\sqrt{9+x^2}}{x} +C[/m]