Cfrgtkky

Problem

Evaluate $$int_0^{2pi}(t-sin t)(1-cos t)^2{rm d}t.$$

Comment

It's very complicated to compute the integral applying normal method. I obtain the result resorting to the skillful formula

$$int_0^{2pi}xf(cos x){rm d}x=piint_0^{2pi}f(sin x){rm d}x,$$
where $f(x) in C[-1,1].$

begin{align*}
require{begingroup}
begingroup
newcommand{dd}{;{rm d}}int_0^{2pi} (t-sin t)(1-cos t)^2 dd t
&= int_0^{2pi} t(1-cos t)^2 dd t - int_0^{2pi} sin t(1-cos t)^2 dd t \
&= piint_0^{2pi} (1-sin t)^2 dd t - int_0^{2pi} (1-cos t)^2 dd (1-cos t) \
&= piint_0^{2pi} left(frac32-frac12cos2t-2sin tright) dd t - left[frac13(1-cos t)^3right]_0^{2pi}\
&= pileft[frac32t-frac14sin2t+2cos tright]_0^{2pi}\
&= 3pi^2
endgroup
end{align*}

But any other solution?

With substitution $t=pi+x$ we see that
$$I=int_0^{2pi}(t-sin t)(1-cos t)^2{rm d}t=int_{-pi}^{pi}(pi+x+sin x)(1+cos x)^2{rm d}x.$$
the part $(x+sin x)(1+cos x)^2$ is an odd function so it's integral over $[-pi,pi]$ is zero, then
$$I=piint_{-pi}^{pi}(1+cos x)^2{rm d}x=piint_{-pi}^{pi}dfrac32+2cos x+dfrac12cos2x{rm d}x=color{blue}{3pi^2}$$

Notice that your integral is in fact an area integral of the function $(x,y) mapsto x$ over one arch of a cycloid given by
$$begin{cases} x = t - sin t \ y = 1-cos tend{cases}$$
Indeed, if we assume that the cycloid is explicitly given by $y = f(x)$, we have
begin{align}
int_{text{cycloid}} x ,dA &= int_0^{2pi} int_0^{f(x)} x,dy,dx\
&= int_0^{2pi} xf(x),dx \
&= begin{bmatrix} x = t-sin t \ f(x) = 1-cos t \ dx = (1-cos t),dtend{bmatrix}\
&= int_0^{2pi} (t-sin t)(1-cos t)^2,dt
end{align}

Now recall the formula for the $x$-coordinate of the centroid:
$$frac1A int_{text{cycloid}} x ,dA = xtext{-coordinate of the centroid} = pi$$

since centroid is clearly at $x = pi$ by symmetry.

Using the same metrod as above area of the cycloid is $$A = int_{text{cycloid}}dA = int_0^{2pi} (1-cos t)^2 ,dt = 3pi$$

this integral being a lot easier than the original one.

It follows $$int_{text{cycloid}} x ,dA = Api = 3pi^2$$

The integral
$$
int_0^{2pi}sin t(1-cos t)^2,dt=Bigl[frac{(1-cos t)^3}{3}Bigr]_0^{2pi}=0
$$
is immediate. Thus we can concentrate on
$$
int_0^{2pi}t(1-cos t)^2,dt=[dots t=2u dots]=
16int_0^{pi}usin^4u,du=int_0^pi u(e^{iu}-e^{-iu})^4,du
$$
For integer $a$, we have
$$
int_0^pi ue^{2iau},du=Bigl[frac{ue^{2iau}}{2ia}Bigr]_0^pi-frac{1}{2ia}int_0^pi e^{2iau},du=frac{pi e^{2iapi}}{2ia}+frac{1}{4a^2}Bigl[e^{2iau}Bigr]_0^pi=frac{pi}{2ia}
$$
Since $(e^{iu}-e^{-iu})^4=e^{4iu}-4e^{2iu}+6-4e^{-2iu}+e^{-4iu}$ we see that
$$
int_0^pi usin^4u,du=int_0^pi 6u,du=3pi^2
$$

begin{align}
&int_0^{2pi}(t-sin t)(1-cos t)^2{rm d}t=int_0^{2pi}t+t{cos }^2 t-2tcos t-sin t -sin t{cos }^2 t+sin 2t{rm d}t\
&=left[frac{1}{2}t^2+frac{1}{2}t^2+frac{1}{4}tsin 2t-frac{1}{4}t^2+frac{1}{8}cos 2t-2tsin t-2cos t+cos t+frac{1}{3}{cos }^3t-frac{1}{2}cos 2tright]_0^{2pi} \
&=left[frac{3}{4}t^2+frac{1}{4}tsin 2t-frac{3}{8}cos 2t-2tsin t-cos t+frac{1}{3}{cos }^3 tright]_0^{2pi}\
&=3pi ^2
end{align}