A calculus problem by Kunal Gupta

$\begin{aligned} f(x) & = \sum_{n=0}^\infty \frac {{\color{#3D99F6}2^{2n-1}}(-1)^n x^n \ln^{2n}(x^2+1)\sin^n x}{(2n)!} \\ & = {\color{#3D99F6}\frac 12} \sum_{n=0}^\infty \frac {(-1)^n {\color{#3D99F6}2^{2n}} x^n \ln^{2n}(x^2+1)\sin^n x}{(2n)!} \\ & = \frac 12 \sum_{n=0}^\infty \frac {(-1)^n {\color{#3D99F6} \left( 2 \sqrt {x \sin x} \ln (x^2+1)\right)}^{2n}}{(2n)!} \\ & = \frac 12 \cos {\color{#3D99F6} \left( 2 \sqrt {x \sin x} \ln (x^2+1)\right)} \end{aligned}$

Then, we have:

$\begin{aligned} \implies \left \lfloor 10^7 f \left(\frac \pi 2 \right) \right \rfloor & = \left \lfloor \frac {10^7}2 \cos \left( 2 \sqrt {\frac \pi 2 \sin \frac \pi 2} \ln \left(\frac {\pi ^2}4 +1 \right) \right) \right \rfloor \\ & = \lfloor -4998457.80786 \rfloor \\ & = \boxed{-4998458} \end{aligned}$