A quick way to make work of some challenging integrals is the following:
Glasser’s Master Theorem: Let $f(x)$ be a Riemann integrable function over $(-\infty,\,\infty)$ then
\[\displaystyle{\int_{-\infty}^\infty f(x)\,\mathrm dx=\int_{-\infty}^\infty f\left(x-\frac{1}{x}\right)\,\mathrm dx.}\] \[\,\]
The proof is rather short and can be seen in Glasser’s original paper.
Example
Suppose we wish to evaluate
\[J=\int_0^\infty\sin^3\left(x^2+\frac{1}{x^2}\right)\,\mathrm dx.\]We can start by seeing
\[\begin{align*}J &= \frac{1}{2}\int_{-\infty}^\infty\sin^3\left(x^2+\frac{1}{x^2}\right)\,\mathrm dx \\ &=\frac{1}{2}\int_{-\infty}^\infty\sin^3\left(\left(x-\frac{1}{x}\right)^2+2\right)\,\mathrm dx. \end{align*}\]Upon applying Glasser’s Master theorem we have
\[\begin{align*}\int_0^\infty\sin^3\left(x^2+\frac{1}{x^2}\right)\,\mathrm dx &=\frac{1}{2}\int_{-\infty}^\infty\sin^3(x^2+2)\,\mathrm dx\\ &=\int_0^\infty\sin^3(x^2+2)\,\mathrm dx \\ &= \frac{1}{4}\int_0^\infty\left(3\sin(x^2+2)-\sin(3x^2+6) \right )\,\mathrm dx \\ &=\frac{1}{4}\int_0^\infty\left(3\cos(2)\sin(x^2)+3\sin(2)\cos(x^2)-\cos(6)\sin(3x^2)-\sin(6)\cos(3x^2)\right)\,\mathrm dx \\ &=\frac{3}{8}\sqrt{\frac{\pi}{2}}(\sin(2)+\cos(2))-\frac{1}{8}\sqrt{\frac{\pi}{6}}(\sin(6)+\cos(6)) \end{align*}\]where in the last line we’ve made use of the Fresnel integrals
\[\int_0^\infty\sin(x^2)\,\mathrm dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}\]and
\[\int_0^\infty\cos(x^2)\,\mathrm dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}.\]