Question

Помогите пожалуйста!!! ​

Answer 1

$4)\ \ y=2+ln(x^2-1)\ \ ,\ \ \ \ 2\leq x\leq 4\ \ ,\\\\l=\int\limits^a_b\, \sqrt{1+(y'(x))^2}\, dx\\\\\\1+(y'(x))^2=1+\Big(\dfrac{2x}{x^2-1}\Big)^2=1+\dfrac{4x^2}{(x^2-1)^2}=\dfrac{x^4-2x^2+1+2x^2}{(x^2-1)^2}=\\\\\\=\dfrac{x^4+2x^2+1}{(x^2-1)^2}=\dfrac{(x^2+1)^2}{(x^2-1)^2}=\Big(\dfrac{x^2+1}{x^2-1}\Big)^2$

$l=\int\limits^4_2\, \sqrt{\Big(\dfrac{x^2+1}{x^2-1}\Big)^2}\, dx=\int\limits^4_2\, \dfrac{x^2+1}{x^2-1}\, dx=\int\limits^4_2\, \dfrac{(x^2-1)+1+1}{x^2-1}\, dx=\\\\\\=\int\limits^4_2\, \Big(1+\dfrac{2}{x^2-1}\Big)\, dx=\Big(x+2\cdot \dfrac{1}{2}\cdot ln\Big|\, \dfrac{x-1}{x+1}\, \Big|\Big)\Big|_2^4=4+ln\, \dfrac{3}{5}-(2+ln\, \dfrac{1}{3})=\\\\\\=2+ln\, \dfrac{9}{5}$

$5)\ \ \left\{\begin{array}{l}x=5(cost+t\, sint)\\y=5(sint-t\, cost)\end{array}\right\ \ \ ,\ \ \ \dfrac{\pi}{2}\leq t\leq \pi \\\\\\l=\int\limits^{t_1}_{t_1}\, \sqrt{(x'(t))^2+(y'(t))^2}\, dt\\\\\\x'(t)=5(-sint+sint+t\, cost)=5t\, cost\\\\y'(t)=5(cost-(cost-t\, sint))=5t\, sint\\\\(x'(t))^2+(y'(t))^2=25t^2\, cos^2t+25t^2\, sin^2t=25t^2(\underbrace{cos^2t+sin^2t}_{1})=25\, t^2$

$l=\int\limits^{\pi }_{\pi /2}\, \sqrt{25t^2}\, dt=\int\limits^{\pi }_{\pi /2}\, 5\, t\, dt=\dfrac{5}{2}\, t^2\, \Big|_{\pi /2}^{\pi}=\dfrac{5}{2}\cdot \Big(\pi ^2-\dfrac{\pi ^2}{4}\Big)=\dfrac{5}{2}\cdot \dfrac{3\pi ^2}{4}=\dfrac{15\pi ^2}{8}$