Evaluate $\int_0^1\frac{\tan^{-1}ax}{x\sqrt{1-x^2}}\,dx$ Evaluate $$\int_0^1\frac{\tan^{-1}ax}{x\sqrt{1-x^2}}\,dx\,,$$  where $a$ being parameter. I am not able to solve this.
 A: Let $$I(a) = \int_0^{1}\frac{\arctan {ax}}{x\sqrt{1-x^2}}dx$$
Differentiating the integral with respect to $a$,
$$I'(a) = \int_0^{1}\frac{dx}{\sqrt{1-x^2}(1+a^2x^2)}$$
Let, $$u^2 = \frac{1-x^2}{1+a^2x^2} \implies x^2=\frac{1-u^2}{1+a^2u^2} \therefore xdx=-\frac{u(1+a^2)}{(1+a^2u^2)^2}du$$
Therefore, the integral reduces to,
$$I'(a)=\frac{1}{\sqrt{1+a^2}}\int_0^{1}\frac{du}{\sqrt{1-u^2}}$$
$$\therefore I'(a)=\frac{\pi}{2\sqrt{1+a^2}} \implies I(a)=\frac{\pi}{2}\ln|a+\sqrt{1+a^2}|+C$$ where C is an undetermined constant that can be found by putting $a=0$ in the original integral which evaluates to $0$. Therefore, $C=0$ and the integral is 
$$I(a)=\frac{\pi}{2}\ln|a+\sqrt{1+a^2}|$$
A: We will use Taylor series of arctan, so we have :
$$f(a)=\int_0^1\frac{\arctan ax}{x\sqrt{1-x^2}}=\sum_{n=0}^\infty{\frac{(-1)^na^{2n+1}}{2n+1}}\biggl(\int_0^1\frac{x^{2n}}{\sqrt{1-x^2}}\biggr)$$
Then we use integration by substitution for calculate the given integral (by choosing $x=\sin t$) :
 $$\int_0^1\frac{x^{2n}}{\sqrt{1-x^2}}=\int_0^{\pi/2}\sin^{2n}t=W_{2n}$$
And we know calculate wallis'integral, 
$$W_{2n}=\frac{\pi}{2}\frac{(2n)!}{2^{2n}(n!)^2}$$
So we have :
$$\int_0^1\frac{\arctan ax}{x\sqrt{1-x^2}}=\frac{\pi}{2}\sum_{n=0}^\infty{\frac{(-1)^n(2n)!}{4^n(2n+1)(n!)^2}a^{2n+1}}$$
Next step we will use taylor series of arcsin 
(see this link https://fr.wikipedia.org/wiki/S%C3%A9rie_de_Taylor), so we have :
 $$f(a)=\frac{\pi}{2}\frac{\arcsin( {ia})}{{i}}=\frac{\pi}{2}\ln|a+\sqrt{1+a^2}|$$
A: Let $a=\tan s$. Under $x\to\sin x$, one has
\begin{eqnarray}
I(s)&:=&\int_0^1\frac{\tan^{-1}(ax)}{x\sqrt{1-x^2}}\,dx\\
&=&\int_0^{\pi/2}\frac{\tan^{-1}(\tan s\sin x)}{\sin x}\,dx.
\end{eqnarray}
Then
\begin{eqnarray}
I'(s)&=&\int_0^{\pi/2}\frac{\sec^2s}{1+\tan^2s\sin^2 x}\,dx\\
&=&\sec^2s\int_0^{\pi/2}\frac{1}{1+\tan^2s\sin^2 x}\,dx\\
&=&\sec^2s\cdot\frac{\pi}{2}\cos s\\
&=&\frac{\pi}{2}\sec s.
\end{eqnarray}
So
$$ I(s)-I(0)=\frac{\pi}{2}\int_0^s\frac{1}{\cos t}dt=\frac{\pi}2\ln(\sec s+\tan s) $$
or
$$ I=\frac{\pi}{2}\ln(a+\sqrt{1+a^2}).$$
