Can we say anything about the integrals of the following type (whether approximately or exact, but not numerically):
$$ I = \int_a^b \frac{1}{F(x)} \sin{\left(\int_a^x F(u) du\right)} dx$$
I am interested in the case when $F(x) = \sqrt{p_4(x)}h(x)$, where $p_4$ is a polynomial of degree 4 with roots $a,b$, and $h(x)$ is an entire complex function. This means the function is holomorphic on an elliptic curve of sorts. Maybe this is doable using the theory of elliptic integrals, but I am unsure on what to look at.
If you prefer, you can also change $F(x)$ if it allows you to say something about $I$. I am just looking to learn how tackle integrals of this type.
Muchas gracias.
I am going to write it slightly differently for obvious reasons. If we say: $$I=\int_a^b\frac{1}{f(x)}\sin\left(\int_0^xf(u)du\right)dx$$ where $f(x)=\sqrt{p_4(x)}h(x)$ then we can define $F'(x)=f(x)$ and so: $$I=\int_a^b\frac{1}{f(x)}\sin\left(F(x)-F(0)\right)dx$$ $$I=\int_a^b\frac{\sin(F(x))\cos(F(0))-\cos(F(x))\sin(F(0))}{f(x)}dx$$ now by letting $u=F(x)\Rightarrow dx=\frac{du}{f(x)}=\frac{du}{u'}$ and so we have: $$I=\int_{F(a)}^{F(b)}\frac{\sin(u)\cos(F(0))-\cos(u)\sin(F(0))}{u'^2}du$$ maybe you can do more with this if you know something about the form of $u'$