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)




$\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 $k=54$


$f(3p+3)$ can be written as a sum of an even and an odd function:

$$f(3p+3)=A+Bx^2+ f'(3)x$$

where I used the fact that the derivative at $p=0$ gives you the linear term coefficient. The even parts are irrelevant because they integrate to zero.

In principle, this is a Taylor expansion of a quadratic function (of course it has only 3 terms and we don't care about two of them).


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