Integrate $ \int_{0}^{1} \frac{\tan^{-1}(x)}{\sqrt{1-x^2}} dx$ without using Hyperbolic Functions

Question

I originally had to solve this Integral: $$ \int_{0}^{1} \frac{\tan^{-1}(x)}{\sqrt{1-x^2}} dx$$

It was suggested to me that I introduce the parameter $a$ and then try Differentiation Under the integral Sign. I thus rewrote the Integral as $$ I(a)=\int_{0}^{1} \frac{\tan^{-1}(ax)}{\sqrt{1-x^2}} dx$$ $$\Longrightarrow I'(a)= \int_0^1\dfrac{y}{(1+(ay)^2)(\sqrt{1-y^2})}dy$$ I thought that I might try integration-by-parts with $\sqrt{1-y^2}$ in the denominator as the derivative of $\sin^{-1}(y)$. However, I don't know how this would help.

Answer 1

Take $y=\sin\left(u\right) $, we get $$\int_{0}^{\pi/2}\frac{\sin\left(u\right)}{1+a^{2}\sin^{2}\left(u\right)}du=\int_{0}^{\pi/2}\frac{\sin\left(u\right)}{a^{2}+1-a^{2}\cos^{2}\left(u\right)}du. $$ Now put $\cos\left(u\right)=v $, then $$\int_{0}^{1}\frac{1}{a^{2}+1-a^{2}v^{2}}dv=\frac{1}{a^{2}+1}\int_{0}^{1}\frac{1}{1-\frac{a^{2}v^{2}}{a^{2}+1}}dv $$ and finally put $\frac{av}{\sqrt{a^{2}+1}}=t $ to get $$\frac{1}{a\sqrt{a^{2}+1}}\int_{0}^{a/\sqrt{a^{2}+1}}\frac{1}{1-t^{2}}dt=\frac{1}{a\sqrt{a^{2}+1}}\tanh^{-1}\left(\frac{a}{\sqrt{a^{2}+1}}\right)=\frac{1}{a\sqrt{a^{2}+1}}\log\left(\sqrt{a^{2}+1}+a\right) $$ using the identity, for $x<1$ $$\tanh^{-1}\left(x\right)=\frac{1}{2}\log\left(\frac{1+x}{1-x}\right).$$

Answer 2

HINT: You might want to use the fact that: $$\dfrac{1}{\sqrt{1-x^2}}=\dfrac{\partial \arcsin(x)}{\partial x}=-\dfrac{\partial \arccos(x)}{\partial x}$$