I have a problem trying to solve the differential equation: $y'= \sqrt{1+y^2}$
The steps that I follow are: $$ y'= \sqrt{1+y^2} \implies \int\frac{dy}{\sqrt{1+y^2}}=\int dx $$ by defining $ y=\tan{(t)}$ in order to calculate the first integral we get: $$ \frac{dy}{dt}=\frac{d(\tan{(t)})}{dt}=\frac{1}{\cos^2{(t)}}\implies dy=\frac{dt}{\cos^2(t)}$$ making the substitution of change of variables in the $y$ integral we get: $$ \int\frac{dt}{\cos(t)}$$ I know what this integral is equal to but I tried an alternative way to calculate it: I used the fact that the integral: $$ \int\frac{dx}{1-x^2}$$ when making a substitution $x=\sin(t)$ transforms to $$\int\frac{dt}{\cos(t)} $$ However the integral above can be easily calculated by breaking it down to fractions as equal to $$\frac{1}{2}\ln{ \left(\frac{1+x}{1-x} \right)}+C $$ which gives us a result $$\int\frac{dt}{\cos(t)}=\frac{1}{2}\ln{ \left(\frac{1+\sin(t)}{1-\sin(t)} \right)}+C $$ which does not coincide however with the result that I am aware of: $$\int\frac{dt}{\cos(t)}=\ln{(|\tan(x)+\sec(x)|)} +C $$ Can someone help me?
There are one and the same! $\frac {1+\sin t} {1-\sin t}=\frac {(1+\sin t)^{2}} {1-\sin^{2}t}$ and $\frac {1+\sin t} {|\cos t|} =|\tan t +\sec t|$.