Question

решить определённый интеграл

Answer 1

$\int x^4\, (x^2-1)^{-\frac{7}{2}}\, dx=\int \dfrac{x^4\, dx}{\sqrt{(x^2-1)^7}}=\Big[\; x=\dfrac{1}{cost}\; ,\; dx=\dfrac{-sint}{cos^2t}\, dt\; ,\\\\\\(x^2-1)=\dfrac{1}{cos^2x}-1=tg^2x\Big]=\int \dfrac{-sint\cdot dt}{cos^4t\cdot tg^7t\cdot cos^2t}=-\int \dfrac{sint\cdot cos^7t\, dt}{cos^6t\cdot sin^7t}=\\\\\\=-\int \dfrac{cost\, dt}{sin^6t}=-\int \dfrac{d(sint)}{sin^6t}=-\dfrac{(sint)^{-5}}{-5}+C=\dfrac{1}{5\, sin^5t}+C=\\\\\\=\dfrac{1}{5\, sin^5(arccos\frac{1}{x})}+C$

$\int \limits _{\sqrt2}^2x^4\, (x^2-1)^{-\frac{7}{2}}\, dx=\dfrac{1}{5\, sin^5(arccos\frac{1}{x})}\; \Big|_{\sqrt2}^2=\dfrac{1}{5}\cdot \Big(\dfrac{1}{sin^5\frac{\pi}{3}}-\dfrac{1}{sin^5\frac{\pi}{4}}\Big)=\\\\\\=\frac{1}{5}\cdot \Big(\dfrac{2^5}{\sqrt{3^5}}-\dfrac{2^5}{\sqrt{2^5}}\Big)=\dfrac{32}{5}\cdot \dfrac{2\sqrt2-3\sqrt3}{6\sqrt6}=\dfrac{16\cdot (2\sqrt2-3\sqrt3)}{15\sqrt6}$