Note that the integrand is periodic, hence: $$I=\lim_{T\to\infty}\frac1{2T}\int_{-T}^T(\sin(t+1)-1)^4\ dt=\frac1{2\pi}\int_0^{2\pi}(\sin(t+1)-1)^4\ dt\\I=\frac1\pi\int_0^\pi(\cos(t)+1)^4\ dt$$ Binomial expanding and Chebyshev polynomials of the first kind: $$\begin{align}8(c(t)+1)^4&=8c^4(t)+3...

Inspired by one of my previous questions, I have found that if $y_0=0$ and $y_{n+1}=\sqrt{2+y_n}$, then $$y_n=2\cos(\pi/2^n)$$ From there, it's easy to see from Taylor expansions that $$y_n=2-\frac{\pi^2}{4^n}+\mathcal O(16^{-n})$$ Does it hold true that if $y_0=0$ and $y_{n+1}=\sqrt{a+y_n...

There are a lot of possible answers. For example, $$\int_{0}^{1} \frac{x^3}{3x^2-3x+1} \mathrm{d} x=\frac{1}{2}$$ or $$\int_{0}^{1}\frac{x^5}{5x^4-10x^3+10x^2-5x+1}\mathrm{d}x=\frac{1}{2}$$ are both good examples of how this property can be used. We can use this property to calculate these compli...