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I got stuck trying to find a general formula for the following integral
$$int_0^{pi} t cdotcos^{2n}{left(frac{t}{2}right)} , dt =
4 int_0^{pi/2} t cdotcos^{2n}t ,dt ; , ; text{ for } n in mathbb{N}.$$
It doesn't seem to be listed on Wikipedias Lists of integrals. After playing around for a while I found the following potentially useful identities$$begin{align}
int_0^{pi} t cdotcos^{2n}{left(frac{t}{2}right)} , dt &=
frac{(2n-1)!!}{(2n)!!} pi^2 - 4int_{0}^{pi/2}{int_0^{t} cos^{2n}{tau} , dtau} , dt \
&= frac{(2n-1)!!}{(2n)!!}cdot left( frac{pi^2}{2} - sum_{k=1}^n frac{(2k)!!}{(2k-1)!! , k^2} right) ,,
end{align}$$
where I would consider the last line a solution if one was able to find a closed formula for the sum that is not given in terms of another integral or another horrible function like Wolfram Alpha suggests.

Edits:

• Moved the power closer to $cos$ to avoid confusion.


• Corrected some constants in the identities.


• Emphasized that $n in mathbb{N}$.

The integral probably does not have a closed form in terms of something simpler than the hypergeometric ${}_{3}F_{2}$ unless something amazing happens with the series. I don't think you can do much better than

$$begin{aligned} I_{n} &= frac{(2n-1)!!}{(2n)!!}frac{sqrt{pi},Gamma^{2}(n+1)}{Gamma(n+3/2)Gamma(n+2)}{}_{3}F_{2}left(1, n+1, n+1; n+3/2, n+2; 1right) \
&= frac{{}_{3}F_{2}left(1, n+1, n+1; n+3/2, n+2; 1right)}{(n+1)(n+1/2)} end{aligned}$$

where I have used the double factorial identity

$$ frac{(2n-1)!!}{(2n)!!} = frac{(2n)!}{2^{2n}(n!)^{2}}$$

and Legendre's duplication formula to simplify the gamma functions and factorials. The hypergeometric function reduces to elementary functions in $x^{2}$ when $n$ is an integer; for example, Mathematica gives the $n=1$ and $n=2$ cases as

$$begin{aligned} {}_{3}F_{2}left(1, 2, 2; 5/2, 3; x^{2}right) &= frac{3}{x^{2}}left(frac{arcsin^{2}x}{x^{2}} - 1right) \
{}_{3}F_{2}left(1, 3, 3; 7/2, 4; x^{2}right) &= frac{15}{8x^{2}}left(frac{3arcsin^{2}x}{x^{4}} - frac{3}{x^{2}} - 1right). end{aligned}$$

The linear term makes things more difficult. By making the substitution $x mapsto pi/2 - x$, one can use the beta function and Legendre's duplication formula to show that the related integral has a nice form

$$begin{aligned} int_{0}^{pi/2}4xleft(cos^{2n}x + sin^{2n}xright)mathrm{d}x &= 2piint_{0}^{pi/2}sin^{2n}x,mathrm{d}x = pi,frac{Gamma(1/2)Gamma(1/2+n)}{Gamma(1+n)} \ &= frac{pi^{2}}{2^{2n}}frac{(2n)!}{(n!)^{2}} = pi^{2}frac{(2n-1)!!}{(2n)!!}. end{aligned}$$

Lastly, the tangent half-angle substitution can be used to write the integral in the form

$$ I_{n} = int_{0}^{pi/2}xcos^{2n}x,mathrm{d}x = int_{0}^{infty}frac{tan^{-1}x}{(1+x^{2})^{n+1}},mathrm{d}x $$

but I do not believe this is much easier than the original integral.

NOT A FULL SOLUTION:

Here, complex analysis (if permitted) is your friend (sort of):

begin{equation}
Releft[tcdot e^{-t^n i} right] = t cdot cosleft(t^nright)
end{equation}

Thus,

begin{align}
I = int_{0}^{pi} t cdot cosleft(t^nright) = Releft[int_{0}^{pi} tcdot e^{-t^n i} :dtright]
end{align}

Here let $u = t^{n}i$ to yield:

begin{align}
I &= Releft[int_{0}^{pi} tcdot e^{-t^n i} :dtright] = Releft[int_{0}^{pi^ni} left(frac{u}{i} right)^{frac{1}{n}}cdot e^{-u} cdot frac{du}{left(frac{u}{i} right)^{frac{n - 1}{n}}}right] \
&= Releft[ i^{1 - frac{2}{n}} int_{0}^{pi^ni} u^{frac{2}{n} - 1}e^{-u}:duright] \&= Releft[i^{1 - frac{2}{n}} gammaleft(u^{frac{2}{n}}, pi^niright)right]
end{align}

Where $gamma(a,b)$ is the lower incomplete gamma function.

One more version - in case it is useful for further simplifications.

Use
$$
cos^{2n}left(frac{t}{2}right)=frac1{2^{2 n}}binom{2 n}{n}+frac1{2^{2 n-1}}{sum _{j=1}^nbinom{2 n}{n-j} cos (jt)}
$$
and
$$
int_0^{pi} t cosleft(ntright) , dt=begin{cases}
frac{pi^2}2,&n=0,\
-frac2{(2k-1)^2},&n=2k-1,\
0,&n=2k>0
end{cases}
$$
to obtain
$$int_0^{pi} t cdotcos^{2n}{left(frac{t}{2}right)} , dt=
frac{pi^2}{2^{2 n+1}}binom{2 n}{n}-frac1{2^{2 n-2}}{sum _{k=1}^{left[frac{n+1}2right]} frac1{(2k-1)^2}binom{2 n}{n+2k-1}}.$$
Mathematica handles the last sum as
$$
frac{pi^2}{2^{2 n+1}}binom{2 n}{n}-frac1{2^{2 n-2}} binom{2 n}{n+1} , _5F_4left(frac{1}{2},frac{1}{2},1,frac{1}{2}-frac{n}{2},1-frac{n}{2};frac{3}{2},frac{3}{2},frac{n}{2}+1,frac{n}{2}+frac{3}{2};1right)
$$
(which is no better than the sum itself).