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Určete integrál:

$$\int  {dx \over \sqrt{a^2 - b^2 x^2}}$$

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Zavedeme substituci:
$$\align b x&=az \\ bx&={a\ dz \over b} \\ z&={bx \over a} \endalign$$

$$\int  {dx \over \sqrt{a^2 - b^2 x^2}}=\int  {a\ dz \over b \sqrt{a^2 - a^2 z^2}} =
\int  {dz \over \sqrt{1 -  z^2}} = {1 \over b}\arcsin z = {1 \over b}\arcsin \left({bx\over a}
\right)$$




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