Nachweis der Kreisflächenfunktion

$$\begin{array}{rlrl}A& =4\ast {\int }_0^{1}f(x)dx& =4\ast {\int }_0^1\sqrt{1-x^2}dx& \\ & =4\ast {\int }_0^1\sqrt{1-{\sin }^{2}u}\ast \cos udu& & \\ & =4\ast {\int }_0^{\frac{\pi }{2}}{\cos }^{2}udu& ={\int }_0^{2\pi }{\cos }^{2}xdx& \\ & =\cos x\ast \sin x+\int {\sin }^{2}xdx& & \\ & =\cos x\sin x-\sin x\ast \cos x+\int {\cos }^{2}xdx& & \text{Hilft nicht}\end{array}$$

$u=\cos x$	$v=\sin x$
$u^{\prime }=\sin x$	$v^{\prime }=\cos x$
${\sin }^{2}x=(\sin x{)}^2=\sin x\ast \sin x$

$u=\sin x$	$v=-\cos x$
$u^{\prime }=\cos x$	$v^{\prime }=\sin x$

$$\begin{array}{rl}& \cos x\ast \sin x+{\int }_0^{2\pi }1-{\cos }^{2}xdx\\ & =\cos x\ast \sin x\int 1dx-\int {\cos }^{2}xdx\\ & \\ & \Rightarrow 2\ast \int {\cos }^{2}xdx=\frac{x+\cos x\sin x}{2}\end{array}$$

Probe:

$$\begin{array}{rl}\text{Aufleiten: }& \frac{1}{2}\left(1+\cos x\cos x-\sin x\sin x\right)\\ =& \frac{1}{2}\left({\cos }^2+{\cos }^{2}x\right)\end{array}$$