Integral $\int_0^{\pi/4}\log \tan x \frac{\cos 2x}{1+\alpha^2\sin^2 2x}dx=-\frac{\pi}{4\alpha}\text{arcsinh}\alpha$ Hi I am trying to prove this
$$
I:=\int_0^{\pi/4}\log\left(\tan\left(x\right)\right)\,
\frac{\cos\left(2x\right)}{1+\alpha^{2}\sin^{2}\left(2x\right)}\,{\rm d}x
=-\,\frac{\pi}{4\alpha}\,\text{arcsinh}\left(\alpha\right),\qquad \alpha^2<1.
$$
What an amazing result this is!  I tried to write
$$
I=\int_0^{\pi/4} \log  \sin x\frac{\cos 2x}{1+\alpha^2\sin^2 2x}-\int_0^{\pi/4}\log \cos x \frac{\cos 2x}{1+\alpha^2\sin^2 2x}dx 
$$
and played around enough here to realize it probably isn't the best idea.
Now back to the original integral I, we can possibly change variables $y=\tan x$ and
re-writing the original integral to obtain
$$
\int_0^{\pi/4}\log \tan x \frac{\cos 2x}{1+{\alpha^2}\big(1-\cos^2 (2x)\big)}dx=\int_0^1 \log y \frac{1-y^2}{1+y^2}\frac{1}{1+{\alpha^2}\big(1-(\frac{1-y^2}{1+y^2})^2\big)}\frac{dy}{1+y^2}.
$$
Simplifying this we have
$$
I=\int_0^1\log y \frac{1-y^2}{1+y^2}\frac{(1+y^2)^2}{(1+y^2)^2+4\alpha^2y^2}\frac{dy}{1+y^2}=\int_0^1\log  y \frac{1-y^2}{(1+y^2)^2+4\alpha^2y^2}dy
$$
Another change of variables $y=e^{-t}$ and we have
$$
I=-\int_0^\infty \frac{t(1-e^{-2t})}{(1+e^{-2t})^2+4\alpha^2 e^{-2t}} e^{-t}dt
$$
but this is where I am stuck...How can we calculate I?  Thanks.
 A: My first step is similar to Ron Gordon but then I took a different route. From integration by parts, the given integral can be written as:
$$-\frac{1}{a}\int_0^{\pi/4} \frac{\arctan(\alpha \sin(2x))}{\sin (2x)}\,dx$$
Consider
$$I(a)=\int_0^{\pi/4} \frac{\arctan(a \sin(2x))}{\sin (2x)}\,dx$$
Differentiate both the sides wrt $a$ to obtain:
$$I'(a)=\int_0^{\pi/4} \frac{1}{1+a^2\sin^2(2x)}\,dx$$
Use the substitution $a\sin(2x)=t$ to obtain:
$$I'(a)=\frac{1}{2}\int_0^a \frac{dt}{\sqrt{a^2-t^2}(1+t^2)}$$
Next use the substitution $t=a/y$ to get:
$$I'(a)=\frac{1}{2}\int_1^{\infty} \frac{y}{\sqrt{y^2-1}(a^2+t^2)}\,dy$$
With yet another substitution which is $y^2-1=u^2$,
$$I'(a)=\frac{1}{2}\int_0^{\infty} \frac{du}{u^2+a^2+1}$$
The final integral is trivial, hence:
$$I'(a)=\frac{\pi}{4\sqrt{a^2+1}}$$
Integrate both sides wrt $a$ to get:
$$I(a)=\frac{\pi}{4}\sinh^{-1}a+C$$
It is easy to see that $C=0$, hence with $a=\alpha$,
$$-\frac{1}{a}\int_0^{\pi/4} \frac{\arctan(\alpha \sin(2x))}{\sin (2x)}\,dx=-\frac{\pi}{4\alpha}\sinh^{-1}\alpha$$
$\blacksquare$
A: I use on Ron's results from integrating by parts:  
$$\alpha I(\alpha) = - \int_0^{\pi/4} dx \frac{\arctan{(\alpha \sin{2 x})}}{\sin{2 x}},$$
As an alternative way to complete the problem, use the method differentiating under the integral sign:
$$\frac{d}{d\alpha}(\alpha I(\alpha)) = -\int_{0}^{\pi/4}\frac{dx}{\alpha^2\sin^2{2x}+1}=-\frac{\pi}{4\sqrt{1+\alpha^2}}\\
\implies \alpha I(\alpha) = -\frac{\pi}{4}\int_{0}^{\alpha}\frac{d\tilde\alpha}{\sqrt{1+\tilde\alpha^2}} = -\frac{\pi}{4}\sinh^{-1}{\alpha}.$$
A: Integrate by parts; then you get that
$$I(\alpha) = \left [\frac1{2 \alpha} \arctan{(\alpha \sin{2 x})} \log{(\tan{x})} \right ]_0^{\pi/4} - \int_0^{\pi/4} dx \frac{\arctan{(\alpha \sin{2 x})}}{\alpha \sin{2 x}}$$
The first term on the RHS is zero.  To evaluate the integral, expand the arctan into a Taylor series and get
$$I(\alpha) = -\frac12 \sum_{k=0}^{\infty} \frac{(-1)^k}{2 k+1} \alpha^{2 k} \int_0^{\pi/2} du \, \sin^{2 k}{u} = -\frac{\pi}{4} \sum_{k=0}^{\infty} \frac{(-1)^k}{2 k+1} \binom{2 k}{k} \left (\frac{\alpha}{2} \right )^{2 k}$$
A little manipulation leads us to
$$\alpha I'(\alpha) +I(\alpha) = -\frac{\pi}{4} \sum_{k=0}^{\infty} (-1)^k \binom{2 k}{k} \left (\frac{\alpha}{2} \right )^{2 k} = -\frac{\pi}{4} \frac1{\sqrt{1+\alpha^2}}$$
The LHS is just $[\alpha I(\alpha)]'$, so the solution is
$$I(\alpha) = -\frac{\pi}{4} \frac{\operatorname{arcsinh}(\alpha)}{\alpha} $$
