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

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 then thought 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 understand how this would help. My friend suggested using hyperbolic functions but I don't know anything about them.  Would somebody please be so kind as to show me how to solve this problem? Many, many thanks in advance!

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 $x<1$ $$\tanh^{-1}\left(x\right)=\frac{1}{2}\log\left(\frac{1+x}{1-x}\right).$$
• Sir, as per what I was told, if we take $$f(a) = \int_{0}^{1} \frac{\arctan(ax)}{\sqrt{1-x^2}}dx$$then $$f'(a) = \frac{\ln(a+\sqrt{1+a^2})}{a\sqrt{1+a^2}}$$ Also, how would we then Integrate $f'(a)$? – Ishan Jun 10 '15 at 19:22
• @BetterWorld You can use, if you prefer, the facts that if $x<1$ holds $$\tanh^{-1}(x)=\frac{1}{2}\log\left(\frac{1+x}{1-x}\right).$$ – Marco Cantarini Jun 10 '15 at 19:26
• Thanks Sir. But Sir, please could you explain to me how to get $$f'(a) = \frac{\ln(a+\sqrt{1+a^2})}{a\sqrt{1+a^2}}$$ from $$f(a) = \int_{0}^{1} \frac{\arctan(ax)}{\sqrt{1-x^2}}dx$$ – Ishan Jun 10 '15 at 19:29
• Sir, please could you explain how you got this step: $$\int_{0}^{1}\frac{1}{a^{2}+1-a^{2}v^{2}}dv=\frac{1}{a+1}\int_{0}^{1}\frac{1}{1-\frac{a^{2}v^{2}}{a+1}}dv$$ According to me, it seems to be $a^2+1$ instead of $a+1$, Sir. – Ishan Jun 10 '15 at 19:35
HINT: You might want to use the fact that: $$\dfrac{1}{\sqrt{1-x^2}}=\dfrac{\partial \arcsin(x)}{\partial x}=-\dfrac{\partial \arccos(x)}{\partial x}$$