Show that $\int_{0}^{\pi/2}{\ln[\cos^2(x)+\sin^2(x)\tan^4(x)]\over \sin^2(x)}\mathrm dx=\pi$ Given that:

$$\int_{0}^{\pi/2}{\ln[\cos^2(x)+\sin^2(x)\tan^4(x)]\over \sin^2(x)}\mathrm dx=\pi\tag1$$

$$1-\sin^2(x)+\sin^2(x)\tan^4(x)$$
$$=1+\sin^2(x)[\tan^4(x)-1]$$
$$=1+\sin^2(x)[\tan^2(x)-1][\tan^2(x)+1]$$
$$=1+\tan^2(x)[\tan^2(x)-1]$$
$$=1-\tan^2(x)+\tan^4(x)$$
$$\int_{0}^{\pi/2}{\ln[1-\tan^2(x)+\tan^4(x)]\over \sin^2(x)}\mathrm dx\tag2$$
$t=\tan(x)\implies dt=\sec^2(x) dx$
$$\int_{0}^{\infty}{\ln(1-t^2+t^4)\over t^2}\mathrm dt\tag3$$
$u=t^2\implies du=2tdt$
$${1\over 2}\int_{0}^{\infty}{\ln(1-u+u^2)\over u^{3/2}}\mathrm du\tag4$$
 A: You are almost there. You may notice that $1-t^2+t^4 = \frac{1+t^6}{1+t^2}$ and
$$ \int_{0}^{+\infty}\frac{\log(1+t^2)}{t^2}\,dt = \pi, \qquad \int_{0}^{+\infty}\frac{\log(1+t^6)}{t^2}\,dt = 2\pi \tag{1} $$
are fairly simple to prove. They both depend on
$$ \int_{0}^{+\infty}\frac{\log(1+t)}{t^\alpha}\,dt \stackrel{\text{IBP}}{=} \frac{\pi}{(1-\alpha)\sin(\pi\alpha)}.\tag{2}$$
A: $J=\displaystyle \int_{0}^{\infty}{\ln(1-t^2+t^4)\over t^2}\mathrm dt$
$\begin{align}
J&=\Big[-\dfrac{\ln(1-t^2+t^4)}{t}\Big]_{0}^{+\infty}+\int_{0}^{+\infty}\dfrac{-2+4t^2}{1-t^2+t^4}dt\\
&=\int_{0}^{+\infty}\dfrac{-2+4t^2}{1-t^2+t^4}dt
\end{align}$
Observe that,
$\displaystyle\int_{0}^{+\infty}\dfrac{t^2}{1-t^2+t^4}dt=\int_{0}^{+\infty}\dfrac{1}{1-t^2+t^4}dt$
(perform the change of variable $y=\dfrac{1}{t}$ )
Therefore,
$\begin{align}
J&=\int_{0}^{+\infty}\dfrac{2}{1-t^2+t^4}dt\\
&=\int_{0}^{+\infty}\dfrac{t^2}{1-t^2+t^4}dt+\int_{0}^{+\infty}\dfrac{1}{1-t^2+t^4}dt\\
&=\int_{0}^{+\infty}\left(1+\dfrac{1}{t^2}\right)\dfrac{1}{1+\left(t-\tfrac{1}{t}\right)^2}dt
\end{align}$
Perform the change of variable $y=t-\dfrac{1}{t}$,
$\begin{align}
J&=\int_{-\infty}^{+\infty}\dfrac{1}{1+t^2}dt\\
&=\boxed{\pi}
\end{align}$
