First, it would be good to know that the area bounded by the curve and the x-axis is convergent to begin with.
[tex]\displaystyle\int_{-\infty}^{\ln\pi}\sin(e^x)\,\mathrm dx[/tex]
Let [tex]u=e^x[/tex], so that [tex]\mathrm dx=\dfrac{\mathrm du}u[/tex], and the integral is equivalent to
[tex]\displaystyle\int_{u=0}^{u=\pi}\frac{\sin u}u\,\mathrm du[/tex]
The integrand is continuous everywhere except [tex]u=0[/tex], but that's okay because we have [tex]\lim\limits_{u\to0^+}\frac{\sin u}u=1[/tex]. This means the integral is convergent - great! (Moreover, there's a special function designed to handle this sort of integral, aptly named the "sine integral function".)
Now, to compute the volume. Via the disk method, we have a volume given by the integral
[tex]\displaystyle\pi\int_{-\infty}^{\ln\pi}\sin^2(e^x)\,\mathrm dx[/tex]
By the same substitution as before, we can write this as
[tex]\displaystyle\pi\int_0^\pi\frac{\sin^2u}u\,\mathrm du[/tex]
The half-angle identity for sine allows us to rewrite as
[tex]\displaystyle\pi\int_0^\pi\frac{1-\cos2u}{2u}\,\mathrm du[/tex]
and replacing [tex]v=2u[/tex], [tex]\dfrac{\mathrm dv}2=\mathrm du[/tex], we have
[tex]\displaystyle\frac\pi2\int_0^{2\pi}\frac{1-\cos v}v\,\mathrm dv[/tex]
Like the previous, this require a special function in order to express it in a closed form. You would find that its value is
[tex]\dfrac\pi2(\gamma-\mbox{Ci}(2\pi)+\ln(2\pi))[/tex]
where [tex]\gamma[/tex] is the Euler-Mascheroni constant and [tex]\mbox{Ci}[/tex] denotes the cosine integral function.
