By the given method in the last entry, we integrate k between a and x. Take a=0, obtaining g(x)=x2 (here m=0). Then the solution is
y(x)=∫x0g(t)√1−g(t)2dt.
In this case,
∫x0t2√1−t4dt.
This integral is not possible to compute!
Other example. Take k(x)=ex. Now
∫x0etdt=ex−1.
If we take m=−1, then g(x)=ex and finally f(x)=∫:0xet√1−e2tdt=arcsin(ex).
In order to make sense in the integrand, we are assuming that x<0. The solution is y(x)=arcsin(ex) whose graphic is
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