$\int_{-1}^0 \sin^{2} (\frac{\pi}{2} f(x+1)) dx +\int_{0}^1 \cos^{2} (\frac{\pi}{2} f(x)) dx =?$

Question

It is well known that we can choose smooth $f:\mathbb R \to [0,1]$ with the property $f(x)=0$ for $x\leq 0$ and $f(x)=1$ for all $x\geq 1.$

Let $0< C<1.$ Can we choose smooth $f:\mathbb R \to [0,1]$ with the property $f(x)=0$ for $x\leq 0$ and $f(x)=1$ for all $x\geq 1$ and it satisfies the following property: $$\int_{-1}^0 \sin^{2} (\frac{\pi}{2} f(x+1)) dx +\int_{0}^1 \cos^{2} (\frac{\pi}{2} f(x)) dx =C?$$

Answer 1

Hint:

Substituting $t=x+1$ you will see that the integral will in fact be equal to $1$ rather than some constant C.

Note: What I don't understand is that the domain you defined for the function is all real numbers but you did not define the function in $(0,1)$ so do you find the value of function in $(0,1)$ or anything else? If you want to define a function for that interval then I must tell you that there would be infinite functions satisfying the integral if $C=1$ and no function if $C\neq 1$