what gets done to f(t) in this integral?

**rainer** · December 9th 2009, 06:47 AM

$\arcsin{(t)}=\int\frac{1}{\sqrt{1-t^2}}dt$

Does this mean that...

$\arcsin{(f(t))}=\int\frac{1}{\sqrt{1-f(t)^2}}dt$ ?

Or what happens to f(t) when this integral is evaluated? Does it just stay the same? That would be nice if it did, but I imagine something complicated happens to it.

**chisigma** · December 9th 2009, 07:25 AM

Suppose to start from...

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

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

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

... so that we obtain...

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

Merry Christmas from Italy

$\chi$ $\sigma$

**HallsofIvy** · December 9th 2009, 07:48 AM

Think of it in terms of something simpler. You know that $\int x dx= \frac{1}{2}x^2+ C$ . But $\int f(x) dx$ is NOT equal to $\frac{1}{2} f^2(x)+ C$ for every function f! That would make integration trivial. What you are doing is essentially a substitution. And if you let u= f(x), then you also have to let du= f'(x)dx.

**rainer** · December 9th 2009, 08:07 AM

Originally Posted by chisigma

Suppose to start from...

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

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

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

... so that we obtain...

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

$\chi$ $\sigma$

Does the $f^{2}(t)$ in equation 3 mean "f of t squared" or "second derivative of f(t)"?

**chisigma** · December 9th 2009, 10:38 AM

Is...

$f(t)\cdot f(t) = f^{2} (t)$ (1)

... and...

$\frac{d^{2}f}{dt^{2}} = f^{(2)} (t)$ (2)

Merry Christmas from Italy

$\chi$ $\sigma$

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