Really tricky integration-----double U and trig substitution The definite integral $h(x) = \sin x/(1 + x^2)$ on the closed interval $[-1,1]$
$\tan^2 x + 1 = \sec^2 x$, $x = \tan \theta$
$$x = \tan \theta$$
$$dx = \sec^2 \theta \, d\theta$$
now I have to find $\sin x$ in terms  of $\theta$
I tried working through this but could not get anywhere.
I feel like there is a double $u$ substitution somewhere....

The integral is $0$. Notice
$$ h(-x) = -h(x) $$
Hence $h$ is odd. In particular
$$ \int_{[-1,1]} h(x) = \int_{[-1,1]} \frac{ \sin x }{1 + x^2} = 0 $$
