Question

If $f$ is continuous for any $tϵR$

then for $I_1={\int }_{{\sin }^{2}t}^{1+{\cos }^{2}t}xf\left(x(2-x)\right)dx$ and $I_2={\int }_{{\sin }^{2}t}^{1+{\cos }^{2}t}f\left(x(2-x)\right)dx$
which of the following is not true?

• A. $I_1={\int }_{{\sin }^{2}t}^{1+{\cos }^{2}t}f\left(x(2-x)\right)dx$
• B. $I_1+I_2=2I_2$
• C. $I_1+I_2=2I_1$
• D. None of the above

Answer 1

The correct option is D None of the above

$I_1={\int }_{{\sin }^{2}t}^{1+{\cos }^{2}t}xf\left(x\right(2-x\left)\right)dx$

Now using the formula:-

${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

Here, $a+b=1+{\cos }^{2}t+{\sin }^{2}t=2$

$I_1={\int }_{{\sin }^{2}t}^{1+{\cos }^{2}t}(2-x)f\left(\right(2-x\left)\right(x\left)\right)dx$

$⟹I_1=2{\int }_{{\sin }^{2}t}^{1+{\cos }^{2}t}f\left(x\right(2-x\left)\right)dx-{\int }_{{\sin }^{2}t}^{1+{\cos }^{2}t}xf\left(x\right(2-x\left)\right)dx$

$⟹I_1=2I_2-I_1$

$⟹I_1=I_2$

Hence, answer is option-(D).