Ответ:
Вычисляем интеграл с помощью тригонометрической подстановки.
[tex]\displaystyle \int \frac{\sqrt{4-x^2}}{x^4}\, dx=\Big[\ x=2sint\ ,\ dx=2cost\, dt\ ,\ 4-x^2=4-4sin^2t=\\\\\\=4(1-sin^2t)=4cos^2t\ \Big]=\int \frac{\sqrt{4cos^2t}}{2^4sin^4t}\cdot 2cost\, dt=\int \frac{4cos^2t}{2^4sin^4t}\, dt=\\\\\\=\frac{1}{4} \int ctg^2t\cdot \frac{dt}{sin^2t}=-\frac{1}{4}\int ctg^2t\cdot d(ctgt)=-\frac{1}{4}\cdot \frac{ctg^3t}{3}+C=\\\\\\=-\frac{1}{12}\cdot ctg^3(arcsin\frac{x}{2})+C=-\frac{1}{12}\cdot \Big(\frac{\sqrt{4-x^2}}{x}\Big)^3+C=-\frac{\sqrt{(4-x^2)^3}}{12\, x^3}+C[/tex]
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Ответ:
Вычисляем интеграл с помощью тригонометрической подстановки.
[tex]\displaystyle \int \frac{\sqrt{4-x^2}}{x^4}\, dx=\Big[\ x=2sint\ ,\ dx=2cost\, dt\ ,\ 4-x^2=4-4sin^2t=\\\\\\=4(1-sin^2t)=4cos^2t\ \Big]=\int \frac{\sqrt{4cos^2t}}{2^4sin^4t}\cdot 2cost\, dt=\int \frac{4cos^2t}{2^4sin^4t}\, dt=\\\\\\=\frac{1}{4} \int ctg^2t\cdot \frac{dt}{sin^2t}=-\frac{1}{4}\int ctg^2t\cdot d(ctgt)=-\frac{1}{4}\cdot \frac{ctg^3t}{3}+C=\\\\\\=-\frac{1}{12}\cdot ctg^3(arcsin\frac{x}{2})+C=-\frac{1}{12}\cdot \Big(\frac{\sqrt{4-x^2}}{x}\Big)^3+C=-\frac{\sqrt{(4-x^2)^3}}{12\, x^3}+C[/tex]