I don't have much experience with hyperbolic trig functions... So I don't know how to start solving this. How do I evaluate the following integrals? Any advice, hint or well-thought solution will be appreciated.
$$I_1= \int\sqrt{\tanh(\ln(\sqrt{x}))} dx$$ $$I_2= \int \ln\left(\sqrt{\tanh(\ln(\sqrt{x}))}\right) dx$$
Background: I made these up while solving similar problems from the internet.. I checked WA for the closed form, but I don't know how to arrive at them...
On
Hint: Recall the definition of hyperbolic tangent.
$$ \tanh x = \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{e^{2x} - 1}{e^{2x} + 1} $$
Substitute this directly into the integrand $\sqrt{\tanh (\ln \sqrt x)}$ to simplify.
$$\sqrt{\tanh(\ln(\sqrt{x}))}=\sqrt{\frac{e^{2\ln(\sqrt x)}-1}{e^{2\ln(\sqrt x)}+1}}=\sqrt{\frac{x-1}{x+1}}$$ Now: $$I_1=\int\sqrt{\frac{x-1}{x+1}}dx\\\stackrel{x=\cosh 2t}=4\int\sinh^2tdt=\sinh(2t)-2t+c\\=\sqrt{x^2-1}-\cosh^{-1}x+c$$
$$I_2\stackrel{x=\cosh 2t}=4\int\sinh(t)\ln(\tanh (t))dt\\=4\cosh t\ln(\tanh t)-4\int {\rm csch} t \\=4\cosh t\ln(\tanh t)-4\ln|{\rm csch} t-\coth t|+c\\= \sqrt2 \sqrt{x+1} \ln\frac{x-1}{x+1}-4 \ln\left(-\tanh\left(\frac 14 \cosh^{-1}(x)\right)\right)+c$$