<s>[INST] Integrating $\int \frac{\sin x-\cos x}{(\sin x+\cos x)\sqrt{(\sin x \cos x + \sin^2x\cos^2x)}}\,dx$ I came across a question today...

Integrate $\int \dfrac{\sin x-\cos x}{(\sin x+\cos x)\sqrt{(\sin x \cos x + \sin^2x\cos^2x)}}\,dx$

How to do it? I tried
1. to take $\sin x \cos x =t$ but no result
2. to convert the thing in the square root into $\sin x +\cos x$ so that I could take $\sin x + \cos x = t$ but then something I got is $\int\frac{-2}{t|t+1|\sqrt{t-1}}\,dt$. Now I don't know how to get past through it.
 [/INST] Notice,
\begin{align}
& \qquad \int\frac{\sin x-\cos x}{(\sin x+\cos x)\sqrt{\sin x\cos x+\sin^2 x\cos^2x}}\ dx\\
& \quad=\int\frac{\sin x-\cos x}{(\sin x+\cos x)\sqrt{\frac{(\sin x+\cos x)^4-1}{4}}}\ dx\\
&\quad=2\int\frac{(\sin x-\cos x)dx}{(\sin x +\cos x)\sqrt{(\sin x+\cos x)^4-1}}
\end{align}
let $\sin x+\cos x=t\implies (\cos x-\sin x)\ dx=dt$,
$$=-2\int\frac{dt}{t\sqrt{t^4-1}} \qquad \qquad\qquad\qquad\qquad$$
let $t^4-1=u^2\implies 4t^3\ dt=2u\ du$,
\begin{align}
&=-2\int\frac{udu}{2u(u^2+1)}\\
&=-\int\frac{du}{1+u^2}\\
&=-\tan^{-1}(u)+C\\
&=-\tan^{-1}\left(\sqrt{t^4-1}\right)+C\\
&=-\tan^{-1}\left(2\sqrt{\sin x\cos x+\sin^2x\cos^2 x}\right)+C\\
\end{align}
</s>