Variable in Feynman Integration

Evaluate: $$I=\int_{0}^{\frac{\pi}{2}} \ln(2468^{2} \cos^2x+990^2 \sin^2x) .dx$$
Now, a friend suggested that to evaluate the integral $I$, I rewrite the integral as $$f(y)=\int_{0}^{\frac{\pi}{2}} \ln(y \cos^2x+ \sin^2x) .dx$$ where $y=\dfrac{2468^2}{990^2}$. $$$$But, how can we do this? $$$$Could somebody please explain how this parametrization was done? Many thanks!
$$$$The suggested solution:$$$$

$$f(y) = \int_{0}^{\pi/2} ln( y^{2}cos^{2}x + sin^{2}x)$$

Here- $ \frac{2468}{990} = y$

$$f'(y) = 2y \int_{0}^{\pi/2}\frac{cos^{2}x}{sin^{2}x + y^{2}cos^{2}x}dx$$

$$= 2y \int_{0}^{\pi/2}\frac{dx}{tan^{2}x + y^{2}}$$

$$= 2y \int_{0}^{\pi/2}\frac{sec^{2}x – tan^{2}x }{tan^{2}x + y^{2}}dx$$

$$= 2y . \frac{1}{y} tan^{-1}( \frac{1}{y}) |_{0}^{\infty} -2y\frac{\pi}{2} + y^{2}f'(y)$$

$$f'(y) = \frac{\pi}{1 + y}$$

$$f(y) = \pi ln(1 + y) + c$$

$$ y = 1 , f(1) = 0 , c = -\pi ln2$$

$$ f(y) = \pi log(1 + y) -\pi ln2 $$

$$$$ According to me, the solution to the evaluation of the Integral $I$ should have been this (I’ve presented an argument below in response to Cameron Sir and DR. MV): $$I=\int_{0}^{\frac{\pi}{2}} \ln(a\cos^2x+ b\sin^2x) .dx$$
$$I=\int_{0}^{\frac{\pi}{2}} \ln(y \cos^2x+ \sin^2x) .dx+\int_0^{\pi/2}\ln (b) dx$$
where $y=\dfrac{a}{b}$. Separating into 2 integrals:$$I=f(y)+\int_0^{\pi/2}\ln (b) dx$$ We then evaluate $f(y)$ and find its value to be $k$ (say). $$\Longrightarrow I=k+\int_0^{\pi/2}\ln (b) dx$$ We can evaluate $\int_0^{\pi/2}ln(b)dx$ and add it to $k$ to get the final value of the Integral $I$.

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$$\log (2468 \cos^2x +990\sin^2x)=\log(990)+\log \left(\frac{2468}{990}\cos^2 x+\sin^2 x\right)$$

and evaluate the integral of $\log 990$ separately.

Then, proceed as in the suggested way forward and use $y=2468/990$ to evaluate at the end of the process.

Why make things more complicated on yourself? If you have $y\cos^2 x + z\sin^2 x$, factor out $z$ to get $z\left(\frac{y}{z}\cos^2 x + \sin^2 x\right)$. After using properties of $\log$, you see that the factor of $z$ out front more or less doesn’t matter and what does matter is $\frac{y}{z}$. Moreover, it doesn’t matter what $y$ or $z$ are but their ratio so you might as well just consider the case your friend suggested.