 #1
chwala
Gold Member
 1,983
 257
 Homework Statement:

$$\int \sqrt{4+x^2}
dx$$
 Relevant Equations:
 hyperbolic equations
still typing...checking latex
That doesn't look so promising to me.Ok to my question now; Could we also use;
##x=2\cosh u## instead of ##x=2\sinh u##?
Cheers
I thought it will be other way round...let me try and see what comes out of it...will share later. Cheers mate.That doesn't look so promising to me.
No. The point is to rewrite ##\sqrt{4 + x^2} = \sqrt{4(1+\sinh^2(u))} = 2 \cosh(u)## using the hyperbolic one. If you would have attempted to use ##x = 2 \cosh(u)## instead you would have ended up with ##\sqrt{4 + x^2} = 2\sqrt{1+\cosh^2(u)}## and ##1 + \cosh^2(u)## does not have any particular simplification in terms of the hyperbolic one. If you would attempt to replace ##\cosh^2(u)## by ##\sinh^2(u)## using the hyperbolic one you would instead end up with ##2 + \sinh^2(u)##, which doesn't make you any happier.Ok to my question now; Could we also use;
##x=2\cosh u## instead of ##x=2\sinh u##?
Thanks @Mark44 Let me study on this approach...Another alternative is to use the substitution ##\tan(\theta) = \frac x 2##. This is based on drawing a right triangle with an acute angle ##\theta##, where the base is 2 and the side opposite is ##x##.
With this substitution, the integral ##\int \sqrt{x^2 + 4}~dx## becomes ##4\int \sec^3(\theta)~d\theta##, an integral so wellknown there's a wikipedia article devoted to it.
correct! i just checked...i ended up with,No. The point is to rewrite ##\sqrt{4 + x^2} = \sqrt{4(1+\sinh^2(u))} = 2 \cosh(u)## using the hyperbolic one. If you would have attempted to use ##x = 2 \cosh(u)## instead you would have ended up with ##\sqrt{4 + x^2} = 2\sqrt{1+\cosh^2(u)}## and ##1 + \cosh^2(u)## does not have any particular simplification in terms of the hyperbolic one. If you would attempt to replace ##\cosh^2(u)## by ##\sinh^2(u)## using the hyperbolic one you would instead end up with ##2 + \sinh^2(u)##, which doesn't make you any happier.