<s>[INST] 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.
 [/INST] 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 $a=\alpha$,
$$-\frac{1}{a}\int_0^{\pi/4} \frac{\arctan(\alpha \sin(2x))}{\sin (2x)}\,dx=-\frac{\pi}{4\alpha}\sinh^{-1}\alpha$$
$\blacksquare$
</s>