TITLE: About $\int_0^{\pi/2}\arctan(1-\sin^2 (x) \cos^2 (x))dx = \pi \left( \frac{\pi}{4}-\arctan \sqrt{\frac{\sqrt{2}-1}{2}}\right)$
QUESTION [18 upvotes]: In this question sos440 has mentioned about an integral that he computed:
$$\int_0^{\pi/2}\arctan(1-\sin^2 (x) \cos^2 (x))dx = \pi \left( \frac{\pi}{4}- \arctan \sqrt{\frac{\sqrt{2}-1}{2}}\right)$$
I would really like to know how this can be proved. I tried using differentiation under the integral sign on parmeter $a$:
$$\int_0^{\pi/2}\arctan(1-a\sin^2 (x) \cos^2 (x))dx $$
but it didn't really work. I would be really grateful if sos440 or somebody else can tell me the secret behind discovering this formula.
Thanks!

REPLY [27 votes]: $$
\begin{aligned} 
\int_0^{\pi/2} \arctan\left(1-\cos^2x\sin^2x\right)\,dx & =\int_0^{\pi/2} \left(\frac{\pi}{2}-\arctan\left(\frac{1}{1-\sin^2x\cos^2x}\right)\right)\,dx \\
&=\frac{\pi^2}{4}-\int_0^{\pi/2} \arctan(\sin^2x)\,dx-\int_0^{\pi/2} \arctan(\cos^2x)\,dx \\
&=\frac{\pi^2}{4}-2\int_0^{\pi/2}\arctan(\cos^2x)\,dx \\
\end{aligned} $$
Consider 
$$I(a)=\int_0^{\pi/2} \arctan(a\,\cos^2x)\,dx $$
$$
\begin{aligned}
\Rightarrow I'(a)&=\int_0^{\pi/2} \frac{\cos^2x}{1+a^2\cos^4x}\,dx \\
&= \int_0^{\pi/2} \frac{\sec^2x}{\sec^4x+a^2}\,dx\\
&= \int_0^{\pi/2} \frac{\sec^2x}{(1+\tan^2x)^2+a^2}\,dx\\
&=\int_0^{\infty} \frac{dt}{t^4+2t^2+a^2+1}\,\,\,\,\,\,\,(\tan x=t) \\
\end{aligned} $$ 
With the substitution $t\mapsto \dfrac{\sqrt{a^2+1}}{t} $,
$$I'(a)=\frac{1}{\sqrt{a^2+1}}\int_0^{\infty} \frac{t^2}{t^4+2t^2+a^2+1}\,dt $$
$$ 
\begin{aligned}
\Rightarrow I'(a) &=\frac{1}{2\sqrt{a^2+1}}\int_0^{\infty} \frac{\sqrt{a^2+1}+t^2}{t^4+2t^2+a^2+1}\,dt \\
&=\frac{1}{2\sqrt{a^2+1}}\int_0^{\infty} \frac{1+\frac{\sqrt{1+a^2}}{t^2}}{\left(t-\frac{\sqrt{a^2+1}}{t}\right)^2+2(1+\sqrt{a^2+1})}\,dt\\
&=\frac{1}{2\sqrt{a^2+1}}\int_{-\infty}^{\infty} \frac{dy}{y^2+2(1+\sqrt{a^2+1})}\,\,\,\,\,\,\,\left(t-\frac{\sqrt{a^2+1}}{t}=y\right) \\
&=\frac{\pi}{2\sqrt{2}} \frac{1}{\sqrt{1+a^2}\sqrt{1+\sqrt{a^2+1}}} \\
\end{aligned} $$ 
Integrating back,
$$
\begin{aligned}
I(1)-I(0)=I(1) &=\frac{\pi}{2\sqrt{2}}\int_0^1 \frac{da}{\sqrt{1+a^2}\sqrt{1+\sqrt{a^2+1}}} \\
&= \frac{\pi}{2\sqrt{2}}\int_1^{\sqrt{2}} \frac{dt}{\sqrt{t-1}(t+1)}\,\,\,\,\,\,\,(\sqrt{a^2+1}=t) \\
&=\frac{\pi}{\sqrt{2}}\int_0^{\sqrt{\sqrt{2}-1}} \frac{du}{u^2+2} \,\,\,\,\,\,\,(t-1=u^2) \\
&=\frac{\pi}{2} \left(\arctan\frac{u}{\sqrt{2}}\right|_0^{\sqrt{\sqrt{2}-1}} \\
&=\frac{\pi}{2}\arctan\left(\sqrt{\frac{\sqrt{2}-1}{2}}\right) \\
\end{aligned}$$ 
Hence,
$$\boxed{\displaystyle \int_0^{\pi/2} \arctan\left(1-\cos^2x\sin^2x\right)\,dx=\dfrac{\pi^2}{4}-\pi\arctan\left(\sqrt{\dfrac{\sqrt{2}-1}{2}}\right)}$$