Вычислим следующий определенный интеграл:
\[\int\limits_{-1}^{1}\dfrac{x^2}{\pi\sqrt{1-x^2}}\,\mathrm{d}x.\]
Решение.
Упростим подинтегральное выражение:
\[\dfrac{x^2}{\sqrt{1-x^2}} = \dfrac{1 - (1 - x^2)}{\sqrt{1-x^2}} = \dfrac{1}{\sqrt{1-x^2}} - \dfrac{1-x^2}{\sqrt{1-x^2}} = \\ = \dfrac{1}{\sqrt{1-x^2}} - \dfrac{\left(\sqrt{1-x^2}\right)^2}{\sqrt{1-x^2}} =\dfrac{1}{\sqrt{1-x^2}} - \sqrt{1-x^2}.\]
Найдем интеграл:
\[\int\dfrac{x^2}{\sqrt{1-x^2}}\,\mathrm{d}x = \int\left(\dfrac{1}{\sqrt{1-x^2}} - \sqrt{1-x^2}\right)\,\mathrm{d}x = \\ = \int\dfrac{1}{\sqrt{1-x^2}}\,\mathrm{d}x - \int\sqrt{1-x^2}\,\mathrm{d}x.\]
(1)
\[\int\dfrac{1}{\sqrt{1-x^2}}\,\mathrm{d}x = \arcsin (x) + const.\]
(2)
\(x = \sin u \rightarrow u = \arcsin(x), \, dx = \cos u\,du\)
\[\int\sqrt{1-x^2}\,\mathrm{d}x = \int\sqrt{1-\sin^2 u}\cdot \cos u\,du = \int\cos^2 u\,du = \\ = \int\frac{1}{2}(\cos 2u + 1)du = \frac{1}{2}\int\cos 2u\,du + \frac{1}{2}\int 1\,du = \\ = \frac{1}{2}\cdot\frac{1}{2}\cdot\int\cos 2u\,d(2u) + \frac{1}{2}u + const = \\ = \frac{1}{4}\sin (2u) + \frac{1}{2}u + const = \frac{1}{4}\sin (2\arcsin (x) ) + \frac{1}{2}\arcsin (x) + const.\]
Таким образом, используя (1) и (2):
\[\int\dfrac{x^2}{\sqrt{1-x^2}}\,\mathrm{d}x = \arcsin (x) - \left(\frac{1}{4}\sin (2\arcsin (x) ) + \frac{1}{2}\arcsin (x) \right) + const.\]
\[\int\dfrac{x^2}{\sqrt{1-x^2}}\,\mathrm{d}x = \frac{1}{2}\arcsin (x) - \frac{1}{4}\sin (2\arcsin (x) ) + const.\]
Следовательно, используя формулу Ньютона - Лейбница, получаем
\[\frac{1}{\pi}\int\limits_{-1}^{1}\dfrac{x^2}{\sqrt{1-x^2}}\,\mathrm{d}x = \frac{1}{\pi}\left(\frac{1}{2}\arcsin (x) - \frac{1}{4}\sin(2\arcsin(x))\right)\Bigr|^{1}_{-1} = \\ = \frac{1}{\pi}\left(\frac{1}{2}(\arcsin (1) - \arcsin(-1)) - \frac{1}{4}(\sin(2\arcsin(1)) - \sin(2\arcsin(-1))\right) =\\ = \frac{1}{\pi}\left(\frac{1}{2}(2\cdot\frac{\pi}{2}) - \frac{1}{4}(2\cdot\sin(2\frac{\pi}{2}))\right) = \frac{1}{\pi}\left(\frac{1}{2}\cdot\pi - \frac{1}{4}\cdot 0\right) = \frac{1}{2}.\]
Ответ.
\[\int\limits_{-1}^{1}\dfrac{x^2}{\pi\sqrt{1-x^2}}\,\mathrm{d}x = \frac{1}{2}.\]
