Definite integral question Let $ f(x)$ be a quadratic equation with $f'(3)=3$. If $I=\int_{0}^{\frac{\pi}{3}}t \times \tan(t)dt $ and the value of integral$\int_{3-\pi}^{{3+\pi}}f(x) \times \tan(\frac{x-3}{3})dx $ is equal to $kI$.Then find k.
My attempt:
Put $\frac{x-3}{3}=p$ in below integral
$$\int_{3-\pi}^{{3+\pi}}f(x) \times \tan\left(\frac{x-3}{3}\right)dx=\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}}f(3p+3) \times \tan(p)\times 3\,dp $$
Now I cannot proceed further, $\tan p$ is an odd function but $f(3p+3)$ is not an even function, neither it is odd function. So their product $f(3p+3) \tan (p)$ cannot say even or odd function.
Can someone help me finding $k$.

Put $\frac{x-3}{3}=p$ in below integral
$\int_{3-\pi}^{{3+\pi}}f(x) \times \tan\left(\frac{x-3}{3}\right)dx=\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}}f(3p+3) \times \tan(p)\times 3\ dp $
Let $f(x)=Ax^2+Bx+C\Rightarrow f'(x)=2Ax+B\Rightarrow f'(3)=6A+B=3$(given in the question)
$f(3p+3)=A(3p+3)^2+B(3p+3)+C=9Ap^2+18Ap+9A+3Bp+3B+C$
$=9Ap^2+(18A+3B)p+9A+3B+C=9Ap^2+3(6A+B)p+9A+3B+C$
$=9Ap^2+3(3)p+9A+3B+C=9Ap^2+9p+9A+3B+C$
$\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}}f(3p+3) \times \tan(p)\times 3\ dp $
$=\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}}(9Ap^2+9p+9A+3B+C) 3 \tan(p) dp$
$=\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}}(9Ap^2+9A+3B+C) 3 \tan(p) dp+\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}}(9p)\times 3 \tan(p) dp$
Function in first integral being an odd function will become zero.
$=\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}}(9p)\times 3 \tan(p) dp$
$=27\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}}p\times tan(p) dp$
$=27\times 2\int_{0}^{\frac{\pi}{3}}p\times tan(p) dp$
$=54\int_{0}^{\frac{\pi}{3}}p\times tan(p) dp$ 
Therefore $k=54$
