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 $k$.