Question

Помогите найти определённый интеграл

Answer 1

$int dfrac{sqrt{x^2-1}}{x^4}, dx=Big[; x=dfrac{1}{cost}; ,; dx=dfrac{sint; dt}{cos^2t}; ,; x^2-1=dfrac{1}{cos^2t}-1=tg^2t; Big]=\\\=int dfrac{tgt}{frac{1}{cos^4t}}cdot dfrac{sint; dt}{cos^2t}=int dfrac{sint; cdot ; sint}{frac{1}{cos^2t}}, dt=int sin^2tcdot cos^2t, dt=int (sintcdot cost)^2, dt=\\\=int Big(; dfrac{1}{2}cdot sin2tBig)^2, dt=dfrac{1}{4}int sin^22t, dt=dfrac{1}{4}int dfrac{1-cos4t}{2}, dt=dfrac{1}{8}int (1-cos4t), dt=$

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

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