Question

Помогите найти определённый интеграл

Answer 1

$\int \dfrac{\sqrt{x^2-1}}{x^4}\, dx=\Big[\; x=\dfrac{1}{cost}\; ,\; dx=\dfrac{sint\; dt}{cos^2t}\; ,\; x^2-1=\dfrac{1}{cos^2t}-1=tg^2t\; \Big]=\\\\\\=\int \dfrac{tgt}{\frac{1}{cos^4t}}\cdot \dfrac{sint\; dt}{cos^2t}=\int \dfrac{sint\; \cdot \; sint}{\frac{1}{cos^2t}}\, dt=\int sin^2t\cdot cos^2t\, dt=\int (sint\cdot cost)^2\, dt=\\\\\\=\int \Big(\; \dfrac{1}{2}\cdot sin2t\Big)^2\, dt=\dfrac{1}{4}\int sin^22t\, dt=\dfrac{1}{4}\int \dfrac{1-cos4t}{2}\, dt=\dfrac{1}{8}\int (1-cos4t)\, dt=$

$=\dfrac{1}{8}\cdot \Big(\; t-\dfrac{1}{4}\, sin4t\; \Big)+C=\dfrac{1}{8}\cdot (arccos\dfrac{1}{x}-\dfrac{1}{4}\, sin(4\, arccos\dfrac{1}{x})\Big)+C\; ;\\\\\\\\\int\limits^2_1\, \dfrac{\sqrt{x^2-1}}{x^4}\, dx=\dfrac{1}{8}\cdot (arccos\dfrac{1}{x}-\dfrac{1}{4}\, sin(4\, arccos\dfrac{1}{x})\Big)\Big|_1^2=\\\\\\=\dfrac{1}{8}\cdot \Big(arccos\dfrac{1}{2}-\dfrac{1}{4}\, sin(4\, arccos\dfrac{1}{2})-arccos1+\dfrac{1}{4}\, sin(4\, arccos1)\Big)=$

$=\dfrac{1}8}\cdot \Big(\dfrac{\pi}{3}-\dfrac{1}{4}\, sin\dfrac{4\pi }{3}-0+\dfrac{1}{4}\, sin0\; \Big)=\dfrac{1}{8}\cdot \Big(\dfrac{\pi}{3}+\dfrac{1}{4}\cdot \dfrac{\sqrt3}{2}\; \Big)=\dfrac{1}{8}\cdot \Big (\dfrac{\pi }{3}+\dfrac{\sqrt3}{8}\; \Big)$