Задача №1451
Условие
Найти интеграл \(\int\frac{\sqrt{1-x^2}}{x^2}dx\).
Решение
\[
\int\frac{\sqrt{1-x^2}}{x^2}dx
=\left[\begin{aligned}& x=\sin{t}.\\& dx=\cos{t}dt.\end{aligned}\right]
=\int\frac{\cos{t}\cdot\cos{t}dt}{\sin^2{t}}
=\int\frac{\cos^2{t}dt}{\sin^2{t}}
=\int\frac{1-\sin^2{t}}{\sin^2{t}}dt
=\int\left(\frac{1}{\sin^2{t}}-1\right)dt=\\
=-\ctg{t}-t+C
=-\frac{\cos{t}}{\sin{t}}-t+C
=-\frac{\sqrt{1-\sin^2{t}}}{\sin{t}}-t+C
=-\frac{\sqrt{1-x^2}}{x}-\arcsin{x}+C
\]
Ответ:
\(-\frac{\sqrt{1-x^2}}{x}-\arcsin{x}+C\)