Recently, I found an interesting integral $$I=\int_{\frac{1}{a}}^a \sin \left(x-\frac{1}{x}\right) \sinh \left(x-\frac{1}{x}\right) d x,$$
I used integration by parts but failed, then I tried by letting $x\mapsto \frac1x$ wonderful which transforms the integral into
$$ \begin{aligned} I & =\int_{\frac{1}{a}}^a \sin \left(\frac{1}{x}-x\right) \sinh \left(\frac{1}{x}-x\right) \frac{d x}{-x^2} \\ & =\int_{\frac{1}{a}}^a \frac{1}{x^2} \sin \left(x-\frac{1}{x}\right) \sinh \left(x-\frac{1}{x}\right) d x \end{aligned} $$ Taking their average tells us that $$ \begin{aligned} I & =\frac{1}{2} \int_{\frac{1}{a}}^a\left(1+\frac{1}{x^2}\right) \sin \left(x-\frac{1}{x}\right) \sinh \left(x-\frac{1}{x}\right) d x \\ & =\frac{1}{2} \int_{\frac{1}{a}}^a \sin \left(x-\frac{1}{x}\right) \sinh \left(x-\frac{1}{x}\right) d\left(x-\frac{1}{x}\right) \\ & =\int_{0}^{a-\frac{1}{a}} \sin u \sinh u d u \end{aligned} $$
For fun, we consider two identical indefinite integrals of the last integral.
$$\begin{aligned}\int \sin u \sinh u d u &=\frac{1}{2}\left[\int \sin u \sinh u d u+\int \sin u \sinh u d u\right]\\&=\frac{1}{2}\left[-\cos u \sinh u+\int \cos u \cosh u d u+\sin u \cosh u-\int \cos u \cosh u\right]\\&=\frac{1}{2} \left[\sin u \cosh u -\cos u \sinh u\right] +C\end{aligned}$$
Now plugging back the limits concludes that $$\boxed{I= \frac{1}{2} \left[\sin \left(a-\frac1a\right) \cosh \left(a-\frac1a\right) -\cos \left(a-\frac1a\right) \sinh \left(a-\frac1a\right)\right] }$$
Is there any other method to evaluate the integral?