Integrating $\int_{0}^{a}\sqrt{{\tanh}(a)-{\tanh}(x)}\;dx$

Question

What is the integral

$$\int_{0}^{a}\sqrt{{\tanh}(a)-{\tanh}(x)}\;dx$$

I have tried various substitutions but couldn't get the answer.

The substitution $x = {\tanh}^{-1}y$ has simplified into rational function in y but I couldn't proceed further.

I got

$$\int_{0}^{\tanh(a)}\frac{1}{\sqrt{\tanh(a)-y}}\frac{1}{1-y^{2}} \; .$$

Answer 1

$$I=\int_0^a \sqrt{\tanh a-\tanh x}dx\overset{\tanh x=y}=\int_0^{\tanh a} \sqrt{\tanh a-y}\ \frac{dy}{1-y^2}$$ Let us denote $\tanh a=n$ for now. And our first goal must be to get rid of the square root, thus a substitution of $n-y =t^2$ would fit great. $$I=\int_0^n \frac{\sqrt{n-y}}{1-y^2}dy=\int^0_{\sqrt n} \frac{t}{1-(n-t^2)^2}(-2t\,)dt=\int_0^{\sqrt n} \frac{2t^2}{1-(n-t^2)^2}dt$$ $$=\int_0^\sqrt n \frac{1+n}{1+n-t^2}dt-\int_0^\sqrt n \frac{1-n}{1-n+t^2}dt$$ $$=\frac{1+n}{\sqrt{1+n}}\operatorname{arctanh}\left(\frac{t}{\sqrt{1+n}}\right)\bigg|_0^\sqrt n-\frac{1-n}{\sqrt{1-n}}\arctan\left(\frac{t}{\sqrt{1-n}}\right)\bigg|_0^\sqrt n$$ $$={\sqrt{1+\tanh a}}\cdot \operatorname{arctanh}\left(\sqrt{\frac{{\tanh a}}{1+\tanh a}}\right)-{\sqrt{1-\tanh a}}\cdot \arctan\left(\sqrt{\frac{{\tanh a}}{1-\tanh a}}\right)$$

Answer 2

Hint

Check if this approach is useful. The substitution $$ y = \sqrt{\tanh(a)-\tanh(x)}$$ gives you $$y^2 = \tanh(a)-\tanh(x),$$ and, therefore, \begin{eqnarray} &&2y dy = -(1-\tanh^2(x)) dx\\ &&2ydy = -\left[1-(\tanh(a)-y^2)^2\right] dx\\ &&\frac{2y dy}{(y^2-\tanh(a))^2-1} = dx. \end{eqnarray} Then the integral becomes rational, i.e. \begin{eqnarray}\mathcal I &=& \int \sqrt{\tanh(a)-\tanh(x)}dx= \\ &=&2\int \frac{y^2dy}{{(y^2-\tanh(a))^2-1}}=\\ &=&2\int\frac{y^2dy}{(y^2-\tanh(a)-1)(y^2+1-\tanh(a))}=\\ &=& 2\int \frac{y^2dy}{(y-\sqrt{\tanh(a)+1})(y+\sqrt{\tanh(a)+1})(y^2+1-\tanh(a))}, \end{eqnarray} recalling that $-1\leq \tanh(x) \leq 1$. You can proceed from here with partial fractions.