Question

Вычислить определенный интеграл. ( сделать замену x=sinx) $\int\limits^1_0 { \sqrt{1- x^{2} } - x^{2} \sqrt{1- x^{2} } } \, dx$

Answer 1

$\int _0^1(\sqrt{1-x^2}-x^2\sqrt{1-x^2})dx=\\\\=[\, x=sint,dx=cost\cdot dt,t=arcsinx,\\t_1=arcsin0=0,t_2=arcsin1=\frac{\pi}{2}\, ]=\\\\=\int _0^{\frac{\pi}{2}}(\sqrt{1-sin^2t}-sin^2t\sqrt{1-sin^2t})\cdot cost\cdot dt=\\\\=\int _0^{\frac{\pi}{2}}(\sqrt{cos^2t}-sin^2t\sqrt{cos^2t})\cdot cost\cdot dt=\\\\=\int_0^{\frac{\pi}{2}}cos^2t\cdot dt-\int _0^{\frac{\pi}{2}}sin^2t\cdot cos^2t\cdot dt=\\\\=\int_0^{\frac{\pi}{2}}\frac{1+cos2t}{2}dt-\int _0^{\frac{\pi}{2}}\frac{1}{4}sin^22t\cdot dt=$

$=(\frac{1}{2}t+\frac{1}{4}sin2t)|_0^{\frac{\pi}{2}}-\frac{1}{4}\int _0^{\frac{\pi}{2}}\frac{1-cos4t}{2}dt=\\\\=\frac{\pi}{4}-\frac{1}{8}(t-\frac{1}{4}sin4t)|_0^{\frac{\pi}{2}}=\frac{\pi}{4}-\frac{1}{8}(\frac{\pi}{2}-0)=\frac{3\pi}{16}$

Answer 2

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