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.
 A: 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}
A: It can be computed as a Lobachevsky-type integral (see also [1] and [2] here on MSE): $$\int_0^\infty\frac{\sin x\,dx}{x(1+\cos^2 x)}=\int_0^{\pi/2}\frac{dx}{1+\cos^2 x}\underset{\tan x=y}{\phantom{\big[}\quad=\quad\phantom{\big]}}\int_0^\infty\frac{dy}{2+y^2}=\frac{\pi}{2\sqrt2}.$$
A: Too long for a comment

Introduce a parameter $a$:
$$\bbox[5px,#ffd]{I(a)=\int \frac{\sin ax}{x(1+\cos ^2 x)}dx}$$
$$I'(a)=\int \frac{\cos ax}{(1+\cos ^2 x)}dx$$
$$I''(a)=\int \frac{-x\sin ax}{(1+\cos ^2 x)}dx=-xI(a)+\int I(a)$$
$$I'''(a)=-xI'(a)+I(a)$$
The solution of this differential equation is this.
Note that $c=-x,y=I(a)$ and $x=a$
