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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!

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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 $x<1$ $$\tanh^{-1}\left(x\right)=\frac{1}{2}\log\left(\frac{1+x}{1-x}\right).$$

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  • $\begingroup$ Sir, could you please give a closed form which does not involve Hyperbolic Functions? $\endgroup$ – Ishan Jun 10 '15 at 19:18
  • $\begingroup$ 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 $f'(a)$? $\endgroup$ – Ishan Jun 10 '15 at 19:22
  • $\begingroup$ @BetterWorld You can use, if you prefer, the facts that if $x<1$ holds $$\tanh^{-1}(x)=\frac{1}{2}\log\left(\frac{1+x}{1-x}\right).$$ $\endgroup$ – Marco Cantarini Jun 10 '15 at 19:26
  • $\begingroup$ 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 $$f(a) = \int_{0}^{1} \frac{\arctan(ax)}{\sqrt{1-x^2}}dx$$ $\endgroup$ – Ishan Jun 10 '15 at 19:29
  • $\begingroup$ 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. $\endgroup$ – Ishan Jun 10 '15 at 19:35
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HINT: You might want to use the fact that: $$\dfrac{1}{\sqrt{1-x^2}}=\dfrac{\partial \arcsin(x)}{\partial x}=-\dfrac{\partial \arccos(x)}{\partial x}$$

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