Lets solve a short integral puzzle

We solve for a particular value given two different integrals from this short integral puzzle.
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[@themibo899]

I did it doing the following:
After a u sub where u=x(2-x), we get that the limits of integration are equal for higher and lower bound, hence meaning the integral is zero, and this is true for both integrals (the bounds all become 1-cos^4(t))
therefore, for all values of t, the integrals are equal to zero, and thus, for all values of t, the fraction i_1/i_2 is 0/0, which is where we will use l'hopital.
We will do the following.
As t approaches some a, where a is any real number, then the final fraction is of the form 0/0, hence we can use l'hopital. After some work of applying the extended fundemental theorem of calculus, we see that in the end i_1/i_2 is equivalent to i_1'(t)/i_2'(t) which is equal to 1. Hence, the fraction is 1 :)