Задача 53740 Помогите, пожалуйста,решить определенный...
Условие
Решение
[m] \int^{1}_{0}\frac{(a^{x}-b^{x})^2}{a^{x}\cdot b^{x}}dx=\int^{1}_{0}\frac{a^{2x}-2a^{x}\cdot b^{x}+b^{2x}}{a^{x}\cdot b^{x}}dx=[/m]
[m]=\int^{1}_{0}(\frac{a^{2x}}{a^{x}\cdot b^{x}}-\frac{2a^{x}\cdot b^{x}}{a^{x}\cdot b^{x}}+\frac{b^{2x}}{a^{x}\cdot b^{x}})dx=[/m]
[m]=\int^{1}_{0}(\frac{a^{x}}{ b^{x}}-2+\frac{b^{x}}{a^{x}})dx=\int^{1}_{0}(\frac{a}{b})^{x}dx-2\int^{1}_{0}dx+\int^{1}_{0}(\frac{b}{a})^{x}dx=[/m]
[m]=(\frac{(\frac{a}{b})^{x}}{ln\frac{a}{b}}-2x+\frac{(\frac{b}{x})^{x}}{ln\frac{b}{a}})|^{1}_{0}=\frac{\frac{a}{b}}{ln\frac{a}{b}}-2+\frac{\frac{b}{a}}{ln\frac{b}{a}}[/m]
2.
[i]Тригонометрическая подстановка:[/i]
[m]x=8sint[/m]
тогда
[m]64-x^2=64-(8sint)^2=64-64sin^2t=64(1-sin^2t)=64cos^2t[/m]
и
[m]dx=d(8sint)=(8sint)`dt=8costdt[/m]
пределы интегрирования
если [m]x=4\sqrt{3}[/m] тогда [m]4\sqrt{3}=8sint[/m]
[m] sint=\frac{\sqrt{3}}{2}[/m] ⇒ [m]t=\frac{\pi}{3}[/m]
если [m]x=0[/m] тогда [m]0=8sint[/m]
[m] sint=0[/m] ⇒ [m]t=0[/m]
[m] \int^{4\sqrt{3}}_{0}\frac{dx}{\sqrt{(64-x^2)^3}}=\int^{\frac{\pi}{3}}_{0}\frac{8costdt}{\sqrt{(64cos^2t})^3}=[/m]
[m]=\int^{\frac{\pi}{3}}_{0}\frac{8cost dt}{8^3cos^3t}=\int^{\frac{\pi}{3}}_{0}\frac{dt}{64cos^2t}=\frac{1}{64}(tgt)^{\frac{\pi}{3}}_{0}=[/m]
[m]=\frac{1}{64}(tg \frac{\pi}{3}-tg0)=\frac{\sqrt{3}}{64}[/m]
3.
Интегрирование по частям:
[m] u=ln^2(x+1)[/m] ⇒ [m] du =2 ln(x+1)\cdot \frac{1}{x+1}dx[/m]
[m]dv=(x+1)^2dx[/m] ⇒ [m] v=\int (x+1)^2dx=\int (x+1)^2d(x+1)=\frac{(x+1)^3}{3}[/m]
Тогда
[m]\int ^{2}_{0} (x+1)^2\cdot ln^2(x+1)dx=(ln^{2}(x+1)\cdot \frac{(x+1)^3}{3})|^{2}_{0}-\int ^{2}_{0} \frac{(x+1)^3}{3}2 ln(x+1)\cdot \frac{1}{x+1}dx=[/m]
[m]=ln^23\cdot\frac{3^3}{3}-ln^21\cdot \frac{1}{3}-\frac{2}{3}\int ^{2}_{0}(x+1)^2\cdot ln(x+1)dx[/m] [red]=[/red]
второй раз интегрирование по частям:
[m] u=ln(x+1)[/m] ⇒ [m] du = \frac{1}{x+1}dx[/m]
[m]dv=(x+1)^2dx[/m] ⇒ [m] v=\int (x+1)^2dx=\int (x+1)^2d(x+1)=\frac{(x+1)^3}{3}[/m]
[red]=[/red][m]ln^23\cdot\frac{3^3}{3}-ln^21\cdot \frac{1}{3}-\frac{2}{3}\cdot ((ln(x+1)\cdot \frac{(x+1)^3}{3})|^{2}_{0}-\int^{2}_{0}\frac{(x+1)^3}{3} \cdot \frac{1}{x+1}dx)=[/m]
[m]=9ln^23-\frac{2}{3}\cdot (ln3\cdot \frac{3^3}{3}-0)+\frac{2}{3}\int^{2}_{0}\frac{(x+1)^2}{3}dx=[/m]
[m]=9ln^23-\frac{2}{3}\cdot (9ln3)+\frac{2}{9}\cdot (\frac{(x+1)^3}{3})|^{2}_{0}=[/m]
[m]=9ln^23-6ln3+\frac{2}{3}\cdot (\frac{3^3}{3}-\frac{1}{3})=9ln^23-6ln3+\frac{52}{27}[/m]