Solution
= Solution
The argument $3x^3+2x$ is odd, so its squared <sine> is even. Replacing $x$ by $-x$ interchanges the numerator and denominator inside the <logarithm>, changing that factor's sign. Thus the full integrand is an <odd function>. It is continuous on the symmetric interval because $|x|\leq1/\pi<1$. Therefore
$$
\boxed{\int_{-1/\pi}^{1/\pi}\sin^2(3x^3+2x)
\log\frac{1-x^5}{1+x^5}\,dx=0}.
$$