Интеграл[latex] \int\limits^0_{-1} { \sqrt{1-x^{2} } } \, dx = \frac{ \pi}{4} [/latex]Почему так и каким образом это вычисляется?

integral [-1;0] dx корень(1-x^2) = ..... {x=sin(t);dx=cos(t)dt} ... = integral [3pi/2;2pi] dt cos^2(t) = = integral  [3pi/2;2pi] dt (cos(2t)+1)/2 = = integral  [3pi/2;2pi] dt cos(2t)/2 + integral [3pi/2;2pi] dt 1/2 = =  sin(2t)/4 [3pi/2;2pi] + t/2 [3pi/2;2pi]= = 0 + pi/4 = pi/4

[latex]\int\sqrt{1-x^2}dx\Rightarrow \left|\begin{array}{ccc}x=sint\\dx=cosdt\end{array}\right|\Rightarrow\int\sqrt{1-sin^2t}\cdot costdt\\\\\\=\int\sqrt{cos^2t}\cdot costdt=\int cost\cdot costdt=\int cos^2tdt=(*)\\----------------------\\cos2t=2cos^2t-1\to2cos^2t=cos2t+1\to cos^2t=\frac{1}{2}(cos2t+1)\\--------------------------\\\\(*)=\int\left[\frac{1}{2}(cos2t+1)\right]dt=\frac{1}{2}\int(cos2t+1)dt=\frac{1}{2}\int cos2tdt+\frac{1}{2}\int dt[/latex] [latex]=\frac{1}{2}sin2t\cdot\frac{1}{2}+\frac{1}{2}t=\frac{1}{4}sin2t+\frac{1}{2}t=\frac{1}{4}\cdot2sintcost+\frac{1}{2}t=\frac{1}{2}sintcost+\frac{1}{2}t\\\\=\frac{1}{2}(sintcost+t)=(*)\\-----------------------------\\x=sint\Rightarrow t=arcsinx\\\\sin^2t+cos^2t=1\to cos^2t=1-sin^2t\to cost=\sqrt{1-sin^2t}\\\to cost=\sqrt{1-x^2}\\----------------------------\\\\(*)=\frac{1}{2}(x\sqrt{1-x^2}+arcsinx)[/latex] =========================================================== [latex]\int\limits_{-1}^0\sqrt{1-x^2}dx=\left\frac{1}{2}(x\sqrt{1-x^2}+arcsinx)\right]^0_{-1}\\\\=\frac{1}{2}(0\sqrt{1-0^2}+arcsin0)-\frac{1}{2}(-1\sqrt{1-(-1)^2}+arcsin(-1))\\\\=\frac{1}{2}\cdot0-\frac{1}{2}(-1\sqrt{1-1}-\frac{\pi}{2})=-\frac{1}{2}\cdot(-\frac{\pi}{2})=\frac{\pi}{4}[/latex]