\input /home/pasky/school/fastex/lib.tex

Primitivn� funkci naz�v�me tak� neur�it� integr�lem �i antiderivac�.

Jak� m� uplatn�n�?
%
Nap��klad fyzik chce zjistit, kde bude ��stice za p�l hodiny, aby se na ni t�eba �el pod�vat.
%
Biolog je vyzbrojen Petriho miskou, m� bakterie, o nich v�, v jak�m pom�ru se mno��, ale
zn� jen ten pom�r. Kdy� polo�� na misku 40 bakteri� a jde na ve�e�i, chce v�d�t, kolik
jich tam bude, a� se r�no vr�t�.

Tedy zn�me derivaci a chceme zjistit p�vodn� funkci.

\subchapter{Z�kladn� vlasnosti primitivn� funkce}{
	\definition{Nech� $f$ je definovan� na otev�en�m intervalu $I$.
		�ekneme, �e funkce $F$ je {\bf primitivn� funkc�} k $f$ na $I$,
		jestli�e pro $\forall x \in I$ plat� $F'(x) = f(x)$.
	}

	\example{
		$f(x) = x^2$: $\left\{\eqalign{F(x) &= {x^3\over3}\cr
		                        G(x) &= {x^3\over3}+1}\right\}$ je primitivn� k $f$.
	}

	\theorem{1}{tvar primitivn� funkce}{
		Nech� $F$, $G$ jsou dv� primitivn� funkce k funkci $f$ na otev�en�m
		intervalu $I$. Potom $\existss C \in \real$ tak, �e
		$$F(x) = G(x) + C\qquad \forall x \in I$$

		\proof{
			$$ H(x) = F(x) - G(x)\qquad x \in I $$
			$$ H'(x) = F'(x) - G'(x) = f(x) - f(x) = 0\qquad x \in I $$
			Tedy je $H$ na $I$ konstantn�.

			\qed
		}

		\notes{
		\list{
			\listItem{Ne ka�d� funkce m� primitivn� funkci.
				(Nap�. $f(x) = \sign x$ na $\real$ nem� primitivn� funkci.)

				M�jme $P = \{f, $ kter� maj� primitivn� funkci$\}$.
			}
			\listItem{$f$ spojit� $\Rightarrow$ $f$ m� primitivn� funkci.
				(D�kaz viz pozd�ji --- \ref{Riemann�v integr�l}.)

				$f$ m� primitivn� funkci $\Rightarrow$ $f$ je \ref{Darboux}ovsk�.
				(D�kaz brzy.)
			}
			\listItem{$F(x) = \cases{x^2\sin\left(1/x^2\right)&$x \ne 0$\cr
			                         0                        &$x = 0$}$

				Potom $F'$ existuje na $\real$, ale nen� spojit�.
				Tedy pro $f = F'$ plat� $f \in P - C$ ($f$ m� prim. funkci,
				ale nen� spojit�).
			}
			\listItem{Jestli�e $f$ m� primitivn� funkci $F$ na $I$, potom
				$F$ je spojit� na $I$, nebo� $F$ m� v�ude na $I$ vlastn�
				derivaci.
			}
		}
		}
	}

	\penalty -100
	\scpart{Zna�en�:}{
		$$ \int f(x)\,dx = \textbox{mno�ina v�ech funkc� primitivn�ch k $f$ na $I$} $$
		$$ x \in I: \int f(x)\,dx = F(x) \Longleftrightarrow F' = f \textbox{na} I $$

		P��eme:
		$$ \int f(x)\,dx = F(x) + C\qquad x \in I $$
		$$ \undernote{\int}{\textbox{\eightrm znak}\atop\textbox{\eightrm integr�lu}}
		   \undernote{f(x)}{\textbox{\eightrm integrant}}\,
		   \undernote{dx}{{{\textbox{\eightrm diferenci�l}
		   	\atop\textbox{\eightrm (ozna�uje}}
		   	\atop\textbox{\eightrm integra�n�}}
		   	\atop\textbox{\eightrm prom�nnou)}}
		   =
		   \undernote{F(x)}{{\textbox{\eightrm reprezentant}\atop
		                    \textbox{\eightrm v�ech integra�n�ch}}\atop
		                    \textbox{\eightrm funkc�}} +
		   \undernote{C}{\textbox{\eightrm aditivn� konstanta}\atop
		   		 \scriptstyle \in \real} $$
	}

	\theorem{2}{linearita primitivn� funkce}{
		Nech� $f$ m� primitivn� funkci $F$ na otev�en�m intervalu $I$, $g$ m� primitivn�
		funkci $G$ na $I$, nech� $\alpha,\beta \in \real$. Potom $(\alpha F + \beta G)$
		je primitivn� funkc� k funkci $(\alpha f + \beta g)$.

		\proof{
			$$ (\alpha F + \beta G)' = \alpha F' + \beta G' = \alpha f + \beta g $$
			\qed
		}

		\scpart{D�sledek:}{
			$$ \int f(ax+b)\,dx = {1\over a} F(ax+b)+C\qquad a \ne 0 $$
		}

		\examples{
		\list{
			\listItem{$\displaystyle{\int x^\alpha\,dx = {x^{\alpha+1}\over\alpha+1}+C}$
				\hfill
				\M{\eqalign{\alpha \in \nat \cup \{0\}: x& \in \real\cr
				           \alpha \in \sZ,\ \alpha < -1: x& \in (-\infty,0) \cup (0,\infty)\cr
					   \alpha \in \real \setminus \sZ: x& \in (0,\infty)}}
			}
			\listItem{$\displaystyle{\int {1\over x}\,dx = \log |x| + C
					}$\hfill ${x \in (-\infty,0) \cup (0,\infty)}$
			}
			\listItem{$\displaystyle{\int e^x\,dx = e^x + C}$\hfill ${x \in \real}$}
			\listItem{$\displaystyle{\int \sin x\,dx = -\cos x + C}$\hfill ${x \in \real}$}
			\listItem{$\displaystyle{\int \cos x\,dx = \sin x + C}$\hfill ${x \in \real}$}
			\listItem{$\displaystyle{\int {1\over\cos^2 x}\,dx = \tg x + C}$\hfill ${x \in (-{\pi/2}+k\pi,{\pi/2}+k\pi),\ k \in \sZ}$}
			\listItem{$\displaystyle{\int {1\over\sin^2 x}\,dx = \cotg x + C}$\hfill ${x \in (0+k\pi,\pi+k\pi),\ k \in \sZ}$}
			\listItem{$\displaystyle{\int {1\over1+x^2}\,dx = \arctg x + C}$\hfill ${x \in \real}$}

			\penalty-100

			\listItem{$\displaystyle{\int {1\over\sqrt{1-x^2}}\,dx = \arcsin x + C}$\hfill ${x \in (-1,1)}$}
			\listItem{$\displaystyle{\int {-1\over\sqrt{1-x^2}}\,dx = \arccos x + C}$\hfill ${x \in (-1,1)}$}
		}
		}
	}

	\penalty -100
	\theorem{3}{vztah spojitosti a existence primitivn� funkce}{
		$f$ spojit� na $I$ $\Rightarrow$ $f$ m� primitivn� funkci na $I$.

		\shortproof{�asem.}
	}

	\theorem{4}{vztah darbouxovskosti a existence primitivn� funkce}{
		Nech� $f$ m� primitivn� funkci na $I$. Potom $f$ je darbouxovsk� na $I$.

		\optnote{\shortnote{Funkce je darbouxovsk�, pokud nab�v� na intervalu v�ech mezihodnot.}}

		\proof{
			Nech� $y_1<y_2$, $y_1,y_2 \in f(I)$. Nav�c nech� $f(x_1) = y_1$, $f(x_2) = y_2$, $x_1,x_2\in I$.
			Bez �jmy na obecnosti p�edpokl�dejme, �e $x_1<x_2$. Nech� $z \in (y_1, y_2)$.
			Chci dok�zat existenci $x^* \in (x_1,x_2)$ takov�ho, �e $f(x^*) = z$.

			Nech� $F$ je primitivn� k $f$. Trik: definuji funkci
			$$ H(x) := F(x) - zx\qquad x \in I $$
			$$ H'(x) = f(x) - z\qquad x \in I $$
			Tedy chci dok�zat, �e $H'(x) = 0$ pro n�jak� $x$. To je n�hodou druh�
			podm�nka existence extr�mu.

			$H$ je spojit�, nebo� m� na $I$ vlastn� derivaci. Tedy je $H$ spojit�
			na $[x_1,x_2]$, tedy $H$ nab�v� sv�ho minima na $[x_1,x_2]$.

			$$ H'(x) = f(x_1) - z = y_1 - z < 0 $$
			Tedy $H$ nem� minimum v $x_1$.
			$$ H'(x) = f(x_2) - z = y_2 - z > 0 $$
			Tedy $H$ nem� minimum v $x_2$.

			To znamen�, �e $H$ m� minimum v n�jak�m bod� $x^* \in (x_1,x_2)$.

			V�me, �e $H$ m� vlastn� derivaci v�ude v $(x_1,x_2)$ $\Rightarrow$ $H'(x^*) = 0$
			$\Rightarrow$ $f(x^*) - z = 0$ $\Rightarrow$ $f(x^*) = z$.

			\qed
		}
	}
}

\scpart{Z�sk�v�n� primitivn�ch funkc�}{
	Jak z�skat primitivn� funkce k dal��m funkc�m?

	\list{
		\listItem{{\bf Z linearity:}

			\list{
			\olistItem{$\displaystyle{\int \cos(2x)\,dx = {1\over2}\sin(2x)+C}$
			}
			\olistItem{Obecn�: $\displaystyle{\int f(x)\,dx = F(x)+C\qquad x \in \real\setminus\{0\}}$,

				pak $\displaystyle{\int f(ax)\,dx = {F(ax)\over a}+C}$.
			}
			\olistItem{
				\M{\int a^x\,dx = \int e^{x\log a}\,dx
					= {1\over\log a}\int\log a\cdot e^{x\log a}\,dx = {a^x\over \log a} + C}
				\hfill
				$a \ne 0$, $a \ne 1$, $a > 0$
			}
			}
		}

			\penalty-100

		\listItem{{\bf Z trigonometrick�ch vzorc�:}

			\list{
			\olistItem{$\displaystyle{\int \cos^2x\,dx = \int{1+\cos(2x)\over2}\,dx
				= \int{1\over2}\,dx + \int{\cos(2x)\over2}\,dx = {x\over2} + {\sin(2x)\over4} + C}$
			}
			\olistItem{$\displaystyle{\int \sin^2\,dx = \int{1-\cos2x\over2}\,dx}$ atd.}
			}
		}
		\listItem{{\bf Z obecn�ch metod integrace.}}
	}

	\scpart{Stru�n� pr�vodce tabulkov�mi hodnotami}{
		\startRuledTable
		\M{\int{1\over\cos^2x} dx} & \M{\tg x}\cr
		\M{\int{1\over1+x^2} dx} & \M{\arctg x}\cr
		\M{\int{z^\alpha dx}} & \M{{z^{\alpha+1}\over\alpha+1}}\cr
		\endRuledTable

		\TODO{V�ce, Je���ku, v�ce!}
	}

	\penalty -100
	\theorem{5}{o substituci}{
	\list{
		\listItem{Nech� $F$ je primitivn� k $f$ na $(a,b)$.
			Nech� $\varphi$ je definovan� na $(\alpha,\beta)$,
			$\varphi\colon (a,b) \to (\alpha,\beta)$
			(s~hodnotami v $(\alpha,\beta)$). Nav�c nech�
			existuje $\varphi'(t)$ vlastn� pro ka�d�
			$t \in (\alpha,\beta)$. Potom
			$$ \int f(\varphi(t))\cdot\varphi'(t)\,dt = F(\varphi(t))+C\qquad t \in (\alpha,\beta) $$
		}
		\listItem{Nech� $\varphi$ m� nenulovou vlastn� derivaci na intervalu $(\alpha,\beta)$,
			$\varphi((\alpha,\beta)) = (a,b)$ (tedy se zobrazuje na cel� interval).
			Nech�
			$$ \int f(\varphi(t))\cdot\varphi'(t)\,dt = G(t)+C\qquad t \in (\alpha,\beta) $$
			Potom
			$$ \int f(x)\,dx = G(\varphi^{-1}(x)) + C\qquad x \in (a,b) $$
		}
	}

		\proof{
		\list{
			\listItem{$F(\varphi(t))' \becauseof{\ref{VODSF}}{=} f(\varphi(t))\cdot\varphi'(t)$}
			\listItem{$\varphi'$ je nenulov� a m� primitivn� funkci $\Rightarrow$ $\varphi'$ je darbouxovsk�
					$\Rightarrow$ $\varphi$ nem�n� znam�nko na $I$
					$\Rightarrow$ $\varphi$ je ryze monot�nn�. Tedy:

					$$ G(\varphi^{-1}(x))' = f(\varphi^{-1}(x))\cdot(\varphi^{-1}(x))'
						\becauseof{\ref{VODIF}}{=} G'(\varphi^{-1}(x))
							\cdot{1\over\varphi'(\varphi^{-1}(x))} = $$

					\penalty-1000

					$$ = f(\varphi(\varphi^{-1}(x))\cdot\varphi'(\varphi^{-1}(x))
							\cdot{1\over\varphi'(\varphi^{-1}(x))} = f(x) $$
			}
		}
			\qed
		}
	}

	\scpart{P��klady na substituci:}{
	\list{
		\listItem{$F' = f$ na $(a,b)$.
			$\varphi$ na $(\alpha,\beta)$, existuje $\varphi'$ vlastn� na $(\alpha,\beta)$,
			$\varphi(\alpha,\beta) \to (a,b)$. Pak
			$$ \int f(\varphi(t))\varphi'(t)\,dt = F(\varphi(t)) + C\qquad t \in (\alpha,\beta) $$

			\examples{
			\list{
				\listItem{\M{\int {\arctg x \over 1+x^2}\,dx\qquad x \in (-\infty, \infty)}

					\startRuledTable
					\M{t \in (-\infty, \infty)} & \M{x \in (-\infty, \infty)}\cr

					\M{\eqalign{\varphi(t)  &= \arctg(t)\cr
						    \varphi'(t) &= {1 \over 1 + t^2}}}&
					\M{\eqalign{f(x)        &= x\cr
						    F(x)        &= {x^2 \over 2} + C}}\cr
					\endRuledTable
					$$ \Rightarrow \int {\arctg x \over 1+x^2}\,dx = {(\arctg x)^2\over2}+C $$

					\scpart{Zrychlen� z�pis:}{
						$$ \eqalign{y &= \arctg x\cr
							    dy &= {dx \over 1+x^2}} $$
						$$ \int {\arctg x\over 1+x^2}\,dx = \int y\,dy = y^2/2 + C
							= {(\arctg x)^2\over2}+C $$

						Pozor, u zrychlen�ho z�pisu je t�eba ov��it p�edpoklady!
					}

				}
				\listItem{\M{\int {1 \over x\log x}\,dx\qquad x \in (1,\infty)}

					\startRuledTable
					\M{\eqalign{y &= \log x\cr
						    dy &= {dx \over x}}}\cr
					\endRuledTable
					$$ \int {1 \over x\log x}\,dx = \int {dy\over y} = \log y + C = \log(\log x) + C $$
					($\existss y'(x)$ vlastn� pro $x \in (1,\infty)$)

					\bigskip
				}
				\listItem{\M{\int {e^{\arccos x}\over\sqrt{1-x^2}}\,dx\qquad x \in (-1,1)}

					$$ (\arccos' x = {-1\over\sqrt{1-x^2}})
						= -\int {e^{\arccos x} \over -\sqrt{1-x^2}}\,dx $$
					\startRuledTable
					\M{\eqalign{y &= \arccos x\cr
						    dy &= -{dx \over \sqrt{1-x^2}}}}\cr
					\endRuledTable
					$$ -\int {e^{\arccos x} \over -\sqrt{1-x^2}}\,dx = -\int e^y\,dy = -e^y + C
						= -e^{\arccos x} + C $$
					($\existss y'(x)$ vlastn� pro $x \in (-1,1)$)

					\bigskip
				}

				\listItem{\M{\int \sin^4x\cos^5x\,dx}

					\startRuledTable
					\M{\eqalign{y &= \sin x\cr
						    dy &= \cos x\,dx}}\cr
					\endRuledTable
					$$ \eqalign{\int \sin^4x\cos^5x\,dx &= \int \sin^4x\cos^4x\cos x\,dx\cr
							&= \int\sin^4x(1-\sin^2x)^2\cos x\,dx\cr
							&= \int y^4(1-y^2)^2\,dy\cr
							&= \int y^4(1-2y^2+y^4)\,dy\cr
							&= \int (y^4 - 2y^6 + y^8)\,dy\cr
							&= {y^5\over5} - {2\over7}y^7 + {y^9\over9} + C\cr
							&= {\sin^5x\over5} - {2\over7}\sin^7x + {\sin^9x\over9} + C} $$

					\bigskip
				}
			}
			}
		}

		\listItem{$\varphi:(\alpha,\beta) \to (a,b)$, tedy $\varphi((\alpha,\beta)) = (a,b)$.
			Existuje $\varphi'$ vlastn� na $(\alpha,\beta)$, $\varphi'(t) \ne 0$ na $(\alpha,\beta)$.
			$$ \int f(\varphi(t))\varphi'(t)\,dt = G(t) + C\qquad t \in (\alpha,\beta) $$

			Pak $$ \int f(x)\,dx = G(\varphi^{-1}(x)) + C\qquad x \in (a,b) $$

			\examples{
			\list{
				\listItem{\M{\int \sqrt{9-x^2}\,dx\qquad x \in (-3,3)}

					Nech� $x = 3\sin t$, $(a,b) = (-3,3)$.
					$$ \varphi(t) = 3\sin t,\ \varphi'(t) = 3\cos t
							\qquad (\alpha,\beta)=(-\pi/2,\pi/2) $$
					$$ \varphi((\alpha,\beta)) = 3\sin((-\pi/2,\pi/s)) = (-3,3) $$

					\startRuledTable
					\M{\eqalign{\varphi(t) &= 3\sin t\cr
					            \varphi^{-1}(t) &= \arcsin(x/3)}}\cr
					\endRuledTable

					$$\eqalign{\int f(\varphi(t))\varphi'(t)\,dt &= \int\sqrt{9-9\sin^2t}\cdot3\cos t\,dt\cr
						&= 9\int\sqrt{1-\sin^2t}\cdot\cos t\,dt\cr
						&\becauseof{$\cos t>0$}{=} 9\int cos^2t\,dt\cr
						&= 9\int{1+\cos(2t)\over2}\,dt\cr
						&= 9\left({t\over2} + {\sin(2t)\over4}\right) + C\cr
						&= 9\left({\arcsin(x/3)\over2} + {\sin(2\arcsin(x/3))\over4}\right) + C}$$
					$$x \in (-3,3)$$

					\scpart{Zkr�cen� z�pis:}{
						$$ \int \sqrt{9-x^2}\,dx $$
						\startRuledTable
						\M{\eqalign{x &= 3\sin t\cr
						            dx&= 3\cos t\,dt\cr
							    t &= \arcsin x/3}}\cr
						\endRuledTable
						$$ \eqalign{\int \sqrt{9-x^2}\,dx &= \int \sqrt{9-9\sin^2t}\cdot3\cos t\,dt\cr
						 	&= 9\left({t\over2} + {\sin(2t)\over4}\right) + C\cr
							&= 9\left({\arcsin(x/3)\over2} + {\sin(2\arcsin(x/3))\over4}\right)+C}$$
					}
				}
			}
			}
		}

		\listItem{Substituce vs. lepen� --- kdy je substituce �patn� n�pad: \M{\int {dt\over 1+\sin^2t}}.
			\smallskip

			Tato funkce je definovan� a spojit� na $\real$, tedy m� primitivn� funkci na $\real$.

			Substituce: $x = \tg t$ pro $t \in (-\pi/2,\pi/2)$ nebo $t \in (-\pi/2+k\pi,\pi/2+k\pi)$, $k \in \intgr$.
			$$ t = \arctg x \Longrightarrow dt = {dx\over1+x^2} $$
			$$ 1+\tg\nolimits^2t = {1\over\cos^2t} \Longrightarrow \cos^2t = {1\over1+x^2} $$
			$$ \Longrightarrow \sin^2t = 1-\cos^2t = 1-{1\over1+x^2} = {x^2\over1+x^2} $$

			Tedy:
			$$ \int {dt\over1+\sin^2t} = \int {1\over1+{x^2\over1+x^2}}{dx\over1+x^2} = \int {dx\over1+2x^2} = $$
			$$ = \int {dx\over1+\left(\sqrt{2}x\right)^2} = {\arctg\left(\sqrt{2}x\right)\over\sqrt{2}} + C $$

			\penalty-100

			Plat� tedy:
			$$ \int {dt\over1+\sin^2t} = {1\over\sqrt{2}}\arctg\left(\sqrt{2}x\right) + C = $$
			$$ = {1\over\sqrt{2}}\arctg\left(\sqrt{2}\tg t\right) + C\qquad t \in (-\pi/2,\pi/2) $$

			Nebo m��eme ��ci:
			$$ \int {dt\over1+\sin^2t} = {1\over\sqrt{2}}\arctg\left(\sqrt{2}\tg t\right) + C\qquad
				t \in (-\pi/2+k\pi,\pi/2+k\pi) $$

			M�me probl�m --- funkce je hrub� nespojit�, tedy to nebude hezk� primitivn� funkce.
			Pou�ijeme proto tzv. mechanismus lepen� --- pospojuji jednotliv� kousky funkce t�m,
			�e je posuneme, aby na sebe hezky navazovaly, a limitami dodefinujeme v nedefinovan�ch
			bodech.

			Na $(-\pi/2,\pi/2)$ zvol�m $c_0 = 0$, tedy \M{F_0(t) = {\arctg\left(\sqrt{2}\tg t\right)\over\sqrt{2}}}.
			\smallskip

			Na $(\pi/2,3\pi/2)$ zvol�m $c_1$ tak, aby \M{\lim_{t\to\pi/2_-} F_0(t) = \lim_{t\to\pi/2+} F_0(t)}.
			\smallskip

			$$\lim_{t\to\pi/2_-} F_0(t) = \lim_{t\to\pi/2_-} {\arctg\left(\sqrt{2}\tg t\right)\over\sqrt{2}}
				= {\pi\over2\sqrt{2}} $$
			$$ \lim_{t\to\pi/2_+} F_0(t) = \lim_{t\to\pi/2_+} {\arctg\left(\sqrt{2}\tg t\right)\over\sqrt{2}} + c_1
				= c_1 + {\pi\over2\sqrt{2}} $$

			Tedy $c_1 = \pi/\sqrt{2}$. Obecn� pak:
			$$ F_0(t) = {\arctg\left(\sqrt{2}\tg t\right)\over\sqrt{2}} + k\pi/\sqrt{2}
				\qquad t \in (-\pi/2+k\pi,\pi/2+k\pi),\ k \in \intgr $$

			Tedy $c_k = k\pi/\sqrt{2}$ pro $k \in \intgr$.

			Dodefinujeme $F_0$ v $\pi/2+k\pi$ hodnotou
			$$ \lim_{t\to\pi/2+k\pi} F_0(t) = {\pi\over2\sqrt{2}} + {k\pi\over\sqrt{2}} $$

			Definujme tedy:
			$$ F_0(t) := \cases{\displaystyle{{\arctg\left(\sqrt{2}\tg t\right)\over\sqrt{2}} + k\pi/\sqrt{2}}
						& $t \in (-\pi/2 + k\pi, \pi/2 + k\pi)$ pro $k \in \intgr$\cr
					    \displaystyle{{\pi\over2\sqrt{2}} + {k\pi\over\sqrt{2}}}
					    	& pro $t = \pi/2 + k\pi$ pro $k \in \intgr$} $$
			Potom:
			$$ \int {dt\over1+\sin^2t} = F_0(t) + C\qquad t \in \real,\ C \in \real $$
		}
	}
	}

	\shortdef{Derivace sou�inu}{$(FG)' = F'G + G'F$}

	\theorem{6}{integrace per partes}{
		Nech� $I$ je otev�en� interval a nech� $f$, $g$ jsou spojit� na $I$.
		Nech� $f$ m� primitivn� funkci $F$ a $g$ m� primitivn� funkci $G$ na $I$.
		Potom plat�:
		$$ \int g(x)F(x)\,dx = F(x)G(x) - \int G(x)f(x)\,dx\qquad x \in I $$

		\proof{
			$$(G(x)F(x))' = g(x)F(x) + G(x)f(x)$$
			$$\Rightarrow g(x)F(x) = (G(x)F(x))' - G(x)f(x)$$
			$$\Rightarrow \int g(x)F(x)\,dx = G(x)F(x) - \int G(x)f(x)\,dx$$
			(Ze spojitosti integr�ly existuj�.)
		}

		\example{
			$$ \int \sin x\cdot x\,dx\qquad x \in \real $$
			\startRuledTable
			\M{F(x) = x} & \M{g(x) = \sin x}\cr
			\M{f(x) = 1} & \M{G(x) = -\cos x}\cr
			\endRuledTable

			$$ \int \sin x\cdot x\,dx = -x\cos x - \int 1\cdot(-\cos x)\,dx = $$
			$$ = -x\cos x + \int \cos x\,dx = \sin x - x\cos x + C $$
		}

			\penalty-100


		\scpart{P��klad (skryt� sou�in):}{
			$$ \int \log x\,dx = \int 1\log x\,dx\qquad x \in (0,\infty)  $$
			\startRuledTable
			\M{F(x) = \log x} & \M{g(x) = 1}\cr
			\M{f(x) = 1/x}    & \M{G(x) = x}\cr
			\endRuledTable

			$$ \int 1\log x\,dx = x\log x - \int x{1\over x}\,dx = x\log x - x + C = x(\log x - x) + C $$
		}

		\scpart{P��klad (v�cen�sobn� per partes):}{
			$$ \int e^x\sin x\,dx = I $$
			\startRuledTable
			\M{F(x) = \sin x} & \M{g(x) = e^x}\cr
			\M{f(x) = \cos x} & \M{G(x) = e^x}\cr
			\endRuledTable
			$$ \Rightarrow I = e^x\sin x - \int e^x\cos x\,dx $$
			\startRuledTable
			\M{F(x) = \cos x}  & \M{g(x) = e^x}\cr
			\M{f(x) = -\sin x} & \M{G(x) = e^x}\cr
			\endRuledTable
			$$ \Rightarrow I = e^x\sin x - \left(e^x\cos x - \int e^x(-\sin x)\,dx\right) = $$
			$$ = e^x\sin x - e^x\cos x - \int e^x\sin x\,dx = e^x(\sin x - \cos x) - I $$
			$$ \Longrightarrow\qquad \eqalign{2I &= e^x(\sin x - \cos x) + C\cr
			                                   I &= e^x(\sin x - \cos x)/2 + C}\qquad x \in \real$$
		}

		\scpart{P��klad (formule pon��en�)}{
			$$ I_n := \int {dx \over (1+x^2)^n}\qquad n \in \nat \cup \{0\} $$
			$$ I_0 = x + c $$
			$$ I_1 = \arctg x + c\qquad x \in \real $$
			$$ I_{n+1} = \int {dx \over (1+x^2)^{n+1}} = \int {1 + x^2 - x^2 \over (1+x^2)^{n+1}} = $$
			$$ = \int {dx \over (1+x^2)^n} - {1\over2}\int{2x\over(1+x^2)^{n+1}}x\,dx = $$
			\startRuledTable
			\M{F(x) = x} & \M{g(x) = {2x \over (1+x^2)^{n+1}}}\cr
			\M{f(x) = 1} & \M{G(x) = {-1 \over n(1+x^2)^n}}\cr
			\endRuledTable
			$$ = I_n + {x \over 2n(1+x^2)^n} - {1\over 2n}\int {dx \over (1+x^2)^n} = $$
			$$ = {x \over 2n(1+x^2)^n} + {2n-1\over2n} I_n $$

			Nap�. pro $n=1$:
			$$ I_2 = \int {dx \over (1+x^2)^2} = {x \over 2(1+x^2)} + {\arctg x\over2} + c $$
		}
	}
}


\scpart{Integrace racion�ln�ch funkc�}{
	M�jme racion�ln� funkci $R(x) = P(x)/Q(x)$, kde $P,Q$ jsou polynomy.
	\shortexample{\M{\int {x^4+1\over x^3 + 5x^2 + 6x - 7}\,dx =\, ?}}

	\medskip
	{\bf Postup p�i integraci racion�ln�ch funkc�:}
	\list{
		\listItem{Je t�eba, aby $\deg P < \deg Q$.
			$$ P(x)/Q(x) = P_1(x) + P_2(x)/Q(x)\qquad \deg P_2 < \deg Q $$

			Toho dos�hneme d�len�m polynom�.

			\penalty-1000

			Tak� jste u� zapomn�li, jak se d�l� mnoho�leny?
			M�jme $$(x^5 + x^4) : (2x^3 + x^2)$$

			Z $Q$ vezmeme nejvy��� mocninu $q = 2x^3$,
			z $P$ takt�: $p = x^5$. Pak provedeme
			$$ r = p/q = {x^5 \over 2x^3} $$
			$$ P := P - rQ $$
			a to opakujeme, dokud $\deg P \ge \deg Q$. K v�sledn�mu $R$
			(sou�tu p�es v�echna $r$) pak p�i�teme zbytek po d�len� $P/Q$.

			Uka�me si cel� postup na rozvlekl�m p��kladu:
			$$ P := x^5 + x^4 $$
			$$ Q := 2x^3 + x^2 \eqno R = 0 $$
			$$ r = {x^5 \over 2x^3} = x^2/2 \eqno R = x^2/2 $$
			$$ P := x^5 + x^4 - x^2/2 \cdot (2x^3 + x^2) = x^5 + x^4 - x^5 - x^4/2 = x^4/2 $$
			$$ r = {x^4/2 \over 2x^3} = x/4 \eqno R = x^2/2 + x/4 $$
			$$ P := x^4/2 - x/4 \cdot (2x^3 + x^2) = x^4/2 - x^4/2 - x^3/4 = -x^3/4 $$
			$$ r = {-x^3/4 \over 2x^3} = 1/8 \eqno R = x^2/2 + x/4 - 1/8 $$
			$$ P := -x^3/4 + 1/8 \cdot (2x^3 + x^2) = -x^3/4 + x^3/4 + x^2/8 = x^2/8 $$
			$$ R = x^2/2 + x/4 - 1/8 + {x^2/8 \over 2x^3 + x^2} $$
		}

		\listItem{Rozklad $Q$ na ko�enov� �initele a irreducibiln� kvadratick� �leny.

			\shortdef{Z�kladn� v�ta algebry}{Ka�d� komplexn� polynom stupn� $n$ m� pr�v� $n$ ko�en�
			($k$-n�sobn� ko�eny po��t�me $k$-kr�t --- ale v $\cmplx$). Pro re�ln� polynom v�ta
			neplat� (nap�. $x^2 + 1$ nem� ko�eny).}

			\shortnote{Polynomy stupn� dva jsou nejvy��� irreducibiln�, nebo� $z \in \cmplx$
			je ko�enem $P$ s re�ln�mi koeficienty $\Longrightarrow$ $\overline{z}$ je ko�en (\DCV).
			Tedy pro $z = \alpha + \beta i$ plat�
			$(x-z)(x-\overline{z}) = x^2 - (z+\overline{z})x + z\overline{z}$
			($z + \overline{z}$ i $z\overline{z}$ $\in \real$).}

			$$ Q(x) = \alpha(x-x_1)^{k_1} \cdots (x-x_2)^{k_l}(x+\alpha_1x + \beta_1)^{s_1}
				\cdots (x^2 + \alpha_nx + \beta_n)^{s_n} $$
		}

		\listItem{Rozklad racion�ln� funkce na parci�ln� zlomky.

			$$ {P(x) \over Q(x)} = {A \over x-x_1} + {B \over (x-x_1)^2}
				+ \cdots + \left(Cx + D \over x^2 + \alpha x + \beta\right)^2 \textnote{apod.} $$

			$A,B,C,D,\ldots$ jsou neur�it� koeficienty, kter� je t�eba spo��tat
			(nap�. pomoc� tzv. cover-up rule nebo fyzik�ln� metody --- prob�r�
			se na cvi�en�).

			Porovn�v�me koeficienty dvou polynom�.
		}

		\listItem{Integrace parci�ln�ch zlomk�.

			Na $(-\infty,x_1)$ nebo na $(x_1,+\infty)$.
			$$ \int{1\over(x-x_1)^k}\,dx =
				\cases{\D{-1\over k^{-1}} {1\over(x-x_1)^{k-1}} + C & $k \ge 2$\cr
				       \log |x-x_1| + C & $k = 1$} $$

			$$ k \in \nat : \int {Cx + D \over \left(x^2+\alpha x + \beta\right)^k}\,dx
				\textbox{irreducibiln�} \Longrightarrow \beta > \alpha^2/4 $$

			$$ = {C \over 2} \int {2x + \alpha - \alpha + 2D/C \over \left(x^2 + \alpha x + \beta\right)^k}\,dx
				= {C \over 2}
					\undernote{\int {2x + \alpha \over \left(x^2 + \alpha x + \beta\right)^k}\,dx}{I_1}
				+ \left(D - {\alpha C \over 2}\right)
					\undernote{\int{dx \over \left(x^2 + \alpha x + \beta\right)^k}}{I_2} $$
			(Konstanty jsou na v�s.)

			$$ I_1 = \cases{\D{-1\over k^{-1}} (x^2 + \alpha x + \beta)^{-k+1} + C & pro $k \ge 2$\cr
				       \log (x^2 + \alpha x + \beta) + C & $k = 1$} \textnote{pro $x \in \real$} $$

			$$ I_2: \textbox{�prava na dvojmoc dvoj�lenu:} x^2 + \alpha x + \beta
				= \left(x + \alpha/2\right)^2 + \undernote{\left(\beta - \alpha^2/4\right)}{>0} $$
			$$ I_2 = \int{dx \over \left(x+\alpha/2\right)^2 + \left(\beta-\alpha^2/4\right)^k}
				= {1 \over \left(\beta - \alpha^2/4\right)^k}
					\cdot
					\int{dx \over \left(\left(x+\alpha/2\over\sqrt{\beta-\alpha^2/4}\right)^2+1\right)^k
					    } $$
			Zvol�m substituci:
			$$ t = {x + \alpha/2 \over \sqrt{\beta-\alpha^2/4}} $$
			$$ \int {dt \over \left(t^2+1\right)^k} \textbox{um�me} $$
		}
	}
}


\penalty-100

\scpart{Integrace trigonometrick�ch funkc�}{
	\shortdef{Polynom}{\M{P(x) = \sum_{j=0}^n a_jx^j,\quad x \in \real, n \in \nat, a_j \in \real, a_n \ne 0}}
	\shortdef{Polynom dvou prom�nn�ch}{\M{P(x,y) = \sum_{i,j=0}^n a_{ij}x^iy^j,\quad a_{ij} \in \real}}
	\shortexample{$x^3 + 2y^3 - 3xy^2 + 14x - 15y^2 - 7$}
	\shortdef{Racion�ln� funkce dvou prom�nn�ch}{\M{R(x,y) = {P(x,y) / Q(x,y)}}}
	
	\medskip
	Zaj�m� n�s integrace funkc� tvaru $R(\sin x, \cos x)$. Nap�.:
	$$ {\sin^2 x \cos x - 4\sin^2 x \over \sin^4 x - 4\sin^2x \cos x - \cos x + 1} $$

	Pou�ijeme substituci:
	\list{
	\listItem{\M{t = \sin x}, je-li $R$ ``lich� v cosinech'', tj. $R(x,-y) = -R(x,y)$.}
	\listItem{\M{t = \cos x}, je-li $R$ ``lich� v sinech'', tj. $R(-x,y) = -R(x,y)$.}
	\listItem{\M{t = \tg x}, je-li $R$ ``sud� v obou'', tj. $R(-x,-y) = R(x,y)$.}
	\listItem{\M{t = \tg (x/2)} (d� se pou��t kdykoliv, ale v�dy vede na racion�ln� funkci).}
	}
}

\bend
