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\rhead{$\mu^{\psi\chi\land\textthreeoldstyle u}$}% mí na psíchí a tri ú :)

{\bf Goniometrické vzorčeky}\\

\noindent
%$$(a^n-b^n) =(a-b)(a^{n-1}+a^{n-2}b+\cdots +ab^{n-2}+b^{n-1})$$
$$1 = \sin^2 x+\cos^2 x$$
$$\sin 2x = 2\sin x\cos x$$
$$\cos 2x = \cos^2 x - \sin^2 x$$
$$\sin (x\pm y) = \sin x\cos y\pm\cos x \sin y$$
$$\cos (x\pm y) = \cos x\cos y\mp\sin x \sin y$$
$$\tan (x\pm y) = \frac{\tan x\pm\tan y}{1\mp\tan x\tan y}$$
$$\sin x\pm\sin y = 2\sin\frac{x\pm y}{2}\cos\frac{x\mp y}{2}$$
$$\cos x+\cos y = 2\cos\frac{x+y}{2}\cos\frac{x-y}{2}$$
$$\cos x-\cos y =-2\sin\frac{x+y}{2}\sin\frac{x-y}{2}$$
$$\sin^2\frac{x}{2} = \frac{1-\cos x}{2} \qquad\qquad \cos^2\frac{x}{2} =\frac{1+\cos x}{2}$$

$$\sin(\arccos x) = \sqrt{1-x^2} \quad\Leftarrow\quad \sin x=\sqrt{1-\cos^2 x}$$
$$\cos(\arcsin x) = \sqrt{1-x^2} \quad\Leftarrow\quad \cos x=\sqrt{1-\sin^2 x}$$
$$\sin(\arctan x) = \frac{x}{\sqrt{1+x^2}} \quad\Leftarrow\quad \sin x=\frac{\tan x}{\sqrt{1+\tan^2 x}}$$
$$\cos(\arctan x) = \frac{1}{\sqrt{1+x^2}} \quad\Leftarrow\quad \cos x=\frac{1}{\sqrt{1+\tan^2 x}}$$
$$\sinh x = \frac{e^x-e^{-x}}{2} \qquad\qquad \cosh x= \frac{e^x+e^{-x}}{2}$$\\

{\bf Limity}\\

$$\lim_{x\to\infty} \frac{\sin x}{x} = 1$$
$$\lim_{x\to\infty} \left(1+\frac{1}{x}\right)^x = e \iff \lim_{x\to 0} \left(1+x\right)^{\frac{1}{x}} = e$$
$$\lim_{x\to 0} \frac{\ln(1+x)}{x} =1 \quad \lim_{x\to 0} \frac{e^x-1}{x} = 1$$

\pagebreak
{\bf Integrály}

$$\int f(x)~dx = F(x)+c$$
$$\int \frac{f^\prime(x)}{f(x)}=\ln |f(x)|$$
$$\int x^n~dx = \frac{x^{n+1}}{n+1}+c$$
$$\int \frac{1}{x}~dx = \ln |x|+c$$
$$\int e^x~dx = e^x+c$$
$$\int a^x~dx = \frac{a^x}{\ln a}+c~{\it (a>0, a\neq 1)}$$
$$\int \frac{1}{1+x^2}~dx=\frac{1}{2}\ln\left |\frac{1+x}{1-x}\right |+c;$$
$$\int \frac{1}{\sqrt{1-x^2}}~dx =\arcsin x+c= -\arccos x+c$$\\

$$\int \sin x~dx=-\cos x+c\quad\int \sinh x~dx=\cosh x+c$$
$$\int \cos x~dx=\sin x+c\qquad\int \cosh x~dx=\sinh x+c$$\\

{\em pre $a>0$}

$$\int \frac{1}{a^2+x^2}~dx=\frac{1}{a}\arctan\frac{x}{a}+c$$
$$\int \frac{1}{a^2-x^2}~dx=\frac{1}{2a}\ln\left |\frac{a+x}{a-x}\right| +c$$
$$\int \frac{1}{\sqrt{a^2+x^2}}~dx=\arcsin\frac{x}{a}+c$$
$$\int \frac{1}{\sqrt{x^2\pm a^2}}~dx=\ln |x+\sqrt{x^2\pm a^2}|+c$$\\

per partes: $\int u^\prime(x)v(x)dx=u(x)v(x)-\int u(x)v^\prime(x)dx$

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