TITLE: Extremely ugly integral $\int_{-\pi}^{\pi} \frac{\operatorname{sign}(x)\arctan(x^2)-|x|}{\sin^2(x)+|\cos(x)|}\,dx$
QUESTION [7 upvotes]: Evaluate: $\DeclareMathOperator{\sign}{sign}$ $$\int_{-\pi}^{\pi} \dfrac{\sign(x)\arctan(x^2)-|x|}{\sin^2(x)+|\cos(x)|}\,dx$$ 

My idea:
\begin{align*}I=\int_{-\pi}^{\pi} \frac{\sign(x)\arctan(x^2)-|x|}{\sin^2(x)+|\cos(x)| }\,dx &=\int_{-\pi}^{\pi} \dfrac{\sign(-x)\arctan((-x)^2)-|-x|}{\sin^2(x)+|\cos(x)|}\,dx\\
&=\int_{-\pi}^{\pi} \frac{-\sign(x)\arctan(x^2)-|x|}{ \sin^2(x)+|\cos(x)| }\,dx.
\end{align*}
So, it means that the integral $$\int_{-\pi}^{\pi} \frac{\sign(x)\arctan(x^2)}{\sin^2(x)+|\cos(x)| }\,dx=0.$$
Therefore, my integral now looks like
$$I=\int_{-\pi}^{\pi} \frac{-|x|}{\sin^2(x)+|\cos(x)|}\,dx,$$
and since my integrand is even function, I have:
$$I=-2\int_{0}^{\pi} \frac{|x|}{ \sin^2(x)+|\cos(x)| }\,dx=-2\int_{0}^{\pi} \frac{x}{ \sin^2(x)+|\cos(x)| }\,dx.$$
And so,
\begin{align*}
\frac{I}{2}&=-\int_{0}^{\pi} \frac{x}{ \sin^2(x)+|\cos(x)| }\,dx\\
&=-\int_{0}^{\pi} \frac{\pi-x}{ \sin^2(\pi-x)+|\cos(\pi-x)|}\,dx\\
&=-\pi\int_{0}^{\pi} \frac{dx}{ \sin^2(x)+|\cos(x)|}+\int_{0}^{\pi} \frac{x}{ \sin^2(x)+|\cos(x)| }\,dx\\
&=-\pi\int_{0}^{\pi} \frac{dx}{ \sin^2(x)+|\cos(x)| }-\frac{I}{2}.
\end{align*}
Hence,
$$I=-\pi\int_{0}^{\pi} \dfrac{dx}{ \sin^2(x)+|\cos(x)| }.$$
That is the end of the road. I tried to eliminate absolute value over the cosine but always get some divergent integral. Is this computation correct and if it is, what next? Bonus question: Where can I find more problems like this?

REPLY [3 votes]: Noting
$$ \int_0^{\pi/2}\frac{1}{\sin^2x+\cos x}dx=\int_0^{\pi/2}\frac{1}{\cos^2x+\sin x}dx=\int_0^{\pi/2}\frac{1}{1-\sin^2x+\sin x}dx $$
and using $u=\tan(\frac{x}{2})$, one has
\begin{eqnarray}
\int_0^{\pi/2}\frac{1}{1-\sin^2x+\sin x}dx&=&\int_0^{\pi/2}\frac{1}{1-(\frac{2\tan(\frac{x}{2})}{1+\tan^2(\frac{x}{2})})^2+\frac{2\tan(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}}dx\\
&=&\int_0^1\frac{1}{1-(\frac{2u}{1+u^2})^2+\frac{2u}{1+u^2}}\frac{2du}{1+u^2}\\
&=&\int_0^1\frac{u^2+1}{u^4+2u^3-2u^2+2u+1}du\\
&=&\int_0^1\frac{u^2+1}{[(u-\phi)^2+\phi][(u+\frac1\phi)^2-\frac1\phi]}du
\end{eqnarray}
Here $\phi=\frac{-1+\sqrt5}{2}$. Noting
$$ \frac{u^2+1}{[(u-\phi)^2+\phi][(u+1/\phi)^2-1/\phi]}=-\frac{\phi}{\sqrt5}\frac{1}{(u-\phi)^2+\phi}+\frac{1}{\sqrt5\phi}\frac{1}{(u+\frac1\phi)^2-\frac1\phi} $$
one has
\begin{eqnarray}
\int_0^{\pi/2}\frac{1}{1-\sin^2x+\sin x}dx
&=&\int_0^1\frac{u^2+1}{[(u-\phi)^2+\phi][(u+\frac1\phi)^2-\frac1\phi]}du\\
&=&-\frac{\phi}{\sqrt5}\int_0^1\frac{1}{(u-\phi)^2+\phi}du+\frac{1}{\sqrt5\phi}\int_0^1\frac{1}{(u+\frac1\phi)^2-\frac1\phi}du\\
&=&-\frac{\sqrt\phi}{\sqrt5}\bigg[\arctan(\frac{u-\phi}{\sqrt\phi})-\text{arctanh}(\frac{1+u\phi}{\sqrt\phi})\bigg]\bigg|_0^1\\
&=&-\frac{\sqrt\phi}{\sqrt5}\bigg[\arctan(\frac{1-\phi}{\sqrt\phi})+\arctan\sqrt\phi-\text{arctanh}(\frac{1}{\phi\sqrt\phi})+\text{arctanh}\sqrt\phi\bigg].
\end{eqnarray}