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**chisigma** Suppose to start from...

$\displaystyle \int \frac{dt}{\sqrt{1-t^{2}}}= \sin^{-1} t + c$ (1)

... and that we want to insert under the integral sign $\displaystyle x=f(t)$ instead of $\displaystyle t$. It will be...

$\displaystyle \frac{dx}{dt} = f^{'} (t) \rightarrow dx= f^{'}(t)\cdot dt$ (2)

... so that we obtain...

$\displaystyle \int \frac{dx}{\sqrt {1-x^{2}}}= \int \frac{f^{'}(t)}{\sqrt{1-f^{2}(t)}}\cdot dt = \sin^{-1} \{f(t)\} + c$ (3)

$\displaystyle \chi$ $\displaystyle \sigma$