Find $\int^\infty_0 \frac{\sin x}{x(1+\cos ^2 x)}dx$ find the value:
$$\int^\infty_0 \frac{\sin x}{x(1+\cos ^2 x)}dx$$
I try to integrate by parts:
$$-\int^\infty_0 \frac{1}{x}d\arctan(\cos x)$$
But it's not run,help me,thank you.

Observe that
\begin{align}
I=&\ \int^\infty_0 \frac{\sin x}{2x(1-\frac{1}{2}\sin^2x)}\ dx = \frac{1}{4}\int^\infty_{-\infty} \frac{\sin x}{x}\sum^\infty_{k=0}\frac{\sin^{2k}x}{2^k}\ dx\\
 =&\ \frac{1}{4}\sum^\infty_{k=0}\frac{1}{2^k}\int^\infty_{-\infty} \frac{\sin^{2k+1}x}{x}\ dx
\end{align}
If you believe that
\begin{align}
\int^\infty_{-\infty} \frac{\sin^{2k+1}x}{x}\ dx = \binom{2k}{k}\frac{\pi}{4^k}
\end{align}
then it follows that
\begin{align}
I=\frac{\pi}{4}\sum^\infty_{k=0}\binom{2k}{k}\frac{1}{8^k} = \frac{\sqrt{2}\pi}{4} = \frac{\pi}{2\sqrt{2}}.
\end{align}
Note that the last line follows from the series expansion
\begin{align}
\frac{1}{\sqrt{1-x}}=\sum^\infty_{k=0}\binom{2k}{k}\frac{x^k}{4^k}.
\end{align}
