Slick way to calculate $\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^T(\sin(t+1)-1)^4\,dt$ by hand? An electrical engineering friend in an introductory signals and systems class asked me for advice on calculating
$$\lim_{T\to\infty}\frac{1}{2T}\int_{-T}^T(\sin(t+1)-1)^4\,dt$$
by hand. I don't see a better way to do this than writing $\sin(t+1)$ using complex exponentials and then doing the binomial expansion. Is there a slicker approach?
 A: 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:
$$\begin{align}8(c(t)+1)^4&=8c^4(t)+32c^3(t)+48c^2(t)+32c(t)+8\\&=c(4t)+8c(3t)+28c(2t)+56c(t)+35\end{align}$$
Thus,
$$\begin{align}8\pi I&=\int_0^\pi c(4t)+8c(3t)+28c(2t)+56c(t)+35\ dt\\&=\frac14s(4t)+\frac83s(3t)+14s(2t)+56s(t)+35t\bigg|_{t=0}^\pi\\&=35\pi\end{align}$$
Thus,

$$I=\frac{35}8$$

A: It is clear we can transform our integral into
$$I=\frac1\pi\int_0^\pi(\cos(t)+1)^4\ dt$$
See @SimplyBeautifulArt's answer for a proof. Once we are here we let $x \to 2x$
$$I=\frac2\pi\int_0^{\pi/2}(\cos(2x)+1)^4\ dx$$
$$I=\frac{2^5}{\pi}\int_0^{\pi/2}\cos^8(x)\ dx$$
By symmetry this is simply
$$I=\frac{2^3}{\pi}\int_0^{2\pi}\cos^8(x)\ dx$$
We now have a fairly standard integral, for which many clever solutions can be found. A fairly general one is to apply the common reduction formulas over repeatedly to find that
$$\int_0^{2\pi}\cos^n(x) dx = \frac{2\pi}{2^n}{n \choose n/2}$$
Applying this closed form here, we find that
$$I=\frac{1}{2^4}{8 \choose 4}= \frac{70}{16}=\frac{35}{8}$$
Of course, we could always use the fact that $\cos^2(x) = 1-\sin^2(x)$ to reduce the power of our integral, or substitute $u=\cos(x)$ or $u=\cos^8(x)$, or any other number of tricks to solve without resorting to the general closed form. I only use the general form because it is relatively well known. See here for a few proofs.
A: take the area of $ (sin(t)+1)^4$, over $2 \pi$, once you expand the $ (sin(t)+1)^4$, sine terms to the odd power give zero area.    so you have to find the area of $sin^4x + 6 Sin^2x+1$. It comes to $70\pi/8$. dividing it by $2\pi$, gives $35/8$.
