Question

Интеграл


$intlimits^0_{-1} { sqrt{1-x^{2} } } , dx = frac{ pi}{4}$
Почему так и каким образом это вычисляется?

Answer 1

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

Answer 2

спасибо!

Answer 3

$intsqrt{1-x^2}dxRightarrow left|begin{array}{ccc}x=sint\dx=cosdtend{array}right|Rightarrowintsqrt{1-sin^2t}cdot costdt\\\=intsqrt{cos^2t}cdot costdt=int costcdot costdt=int cos^2tdt=(*)\----------------------\cos2t=2cos^2t-1to2cos^2t=cos2t+1to cos^2t=frac{1}{2}(cos2t+1)\--------------------------\\(*)=intleft[frac{1}{2}(cos2t+1)right]dt=frac{1}{2}int(cos2t+1)dt=frac{1}{2}int cos2tdt+frac{1}{2}int dt$

$=frac{1}{2}sin2tcdotfrac{1}{2}+frac{1}{2}t=frac{1}{4}sin2t+frac{1}{2}t=frac{1}{4}cdot2sintcost+frac{1}{2}t=frac{1}{2}sintcost+frac{1}{2}t\\=frac{1}{2}(sintcost+t)=(*)\-----------------------------\x=sintRightarrow t=arcsinx\\sin^2t+cos^2t=1to cos^2t=1-sin^2tto cost=sqrt{1-sin^2t}\to cost=sqrt{1-x^2}\----------------------------\\(*)=frac{1}{2}(xsqrt{1-x^2}+arcsinx)$

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$intlimits_{-1}^0sqrt{1-x^2}dx=leftfrac{1}{2}(xsqrt{1-x^2}+arcsinx)right]^0_{-1}\\=frac{1}{2}(0sqrt{1-0^2}+arcsin0)-frac{1}{2}(-1sqrt{1-(-1)^2}+arcsin(-1))\\=frac{1}{2}cdot0-frac{1}{2}(-1sqrt{1-1}-frac{pi}{2})=-frac{1}{2}cdot(-frac{pi}{2})=frac{pi}{4}$