I'm new to this course, I googled similar problems and watched several videos on how to solve similar problems but I'm still not sure how to solve this problem.
As far as I understand the total area of $x^3-2x$ and $C\cdot tg2x$ is equal to $1$.
Am I supposed to find the area of $x^3-2x$ first and then subtract it from $1$?
We want $\int_{-\infty}^\infty f(x)\,dx=1$. Given the piecewise definition of the function, we can split this as $$\int_{-\infty}^{-1}0\,dx+\int_{-1}^0 \left(x^3-2x\right)\,dx+\int_0^{\pi/8}\left(C\cdot tg2x\right)\,dx+\int_{\pi/8}^\infty 0\,dx=1\\ \implies\left.\frac{x^4}{4}-x^2\right|_{-1}^0~~+~~\left.\frac{-C}{2}\log|\cos2x|\right|_{0}^{\pi/8}=1\\ \implies \frac{3}{4}+C\frac{\log2}{4}=1\\ \implies C=\frac{1}{\log2}$$
I hope that clears up any confusion about the process!