I'm having trouble with proving that the double integral $$\int_{-1}^1\int_{x^2}^1 f\Big(\frac{y}{x}\Big)\mathrm{d}y \mathrm{d}x $$ can be simplified to: $$\int_{-1}^1 f(t) \frac{t^2}{2} \mathrm{d}t + \int_{-1}^1 f\Big(\frac{1}{t}\Big) \frac{1}{2} \mathrm{d}t $$
This exercice is part of the section "double integrals in polar coordinates", but I can't see the link. Thank you in advance!
Note that $$\int_{-1}^1dx \int_{x^2}^1 f\Big(\frac{y}{x}\Big)dy= \int_{-1}^1dx \int_{x^2}^{|x|} f\Big(\frac{y}{x}\Big)dy + \int_{-1}^1dx \int_{|x|}^{1} f\Big(\frac{y}{x}\Big)dy.$$ In the first one, let $t=y/x$ then $dxdy=(y/t^2)dtdy$ $$\int_{-1}^1\mathrm{d}x \int_{x^2}^{|x|} f\Big(\frac{y}{x}\Big)dy=\int_{t=-1}^1(1/t^2)f(t)dt \int_{y=0}^{t^2} ydy=\int_{-1}^1 f(t) \frac{t^2}{2} dt.$$ In the second one, let $t=x/y$ then $dxdy=ydtdy$ and $$\int_{x=-1}^1dx \int_{y=|x|}^{1} f\Big(\frac{y}{x}\Big)dy=\int_{t=-1}^1f\Big(\frac{1}{t}\Big)dt \int_{y=0}^{1} ydy=\int_{t=-1}^1f\Big(\frac{1}{t}\Big)\frac{1}{2}dt.$$