Question

For any $tϵR$ and $f$ a continuous function, let $I_1={\int }_{{\sin }^{2}t}^{1+{\cos }^{2}t}xf\left(x(2-x)\right)dx$ and $I_2={\int }_{{\sin }^{2}t}^{1+{\cos }^{2}t}f\left(x(2-x)\right)dx$, then $\frac{I_1}{I_2}$ is equal to

Answer 1

$I_1={\int }_{{\sin }^{2}t}^{1+{\cos }^{2}t}xf\left(x(2-x)\right)dx$
$I_1={\int }_{{\sin }^{2}t}^{1+{\cos }^{2}t}(2-x)f\left((2-x)(2-(2-x))\right)dx$.................$(\because {\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x\left)dx\right)$
$I_1={\int }_{{\sin }^{2}t}^{1+{\cos }^{2}t}(2-x)f\left(x(2-x)\right)dx$
$I_1=2{\int }_{{\sin }^{2}t}^{1+{\cos }^{2}t}f\left(x(2-x)\right)dx-{\int }_{{\sin }^{2}t}^{1+{\cos }^{2}t}xf\left(x(2-x)\right)dx$
$I_1=2I_2-I_1$
$\Rightarrow 2I_1=2I_2\Rightarrow \frac{I_1}{I_2}=1$