Joppy said:
Would you mind sharing your solution? I remember there being a trick with these things but I cannot remember!
Ok
let
$$ cos(x) = \sum_{n=0}^{\infty}C_{n}P_{n}(x)$$
orthogonal
$$ \int_{-1}^{1} cos(x)P_{n}(x)\,dx = \int_{-1}^{1} C_{n}P_{n}(x)P_{n}(x)\,dx$$
$$ \int_{-1}^{1} cos(x)P_{n}(x)\,dx = C_{n}\int_{-1}^{1} P_{n}(x)P_{n}(x)\,dx$$
see that
$$ \int_{-1}^{1} P_{n}(x)P_{n}(x)\,dx= \frac{1}{2^{2n}(n!)^2}\int_{-1}^{1} \frac{d^n}{dx^n}(x^2-1)^n\frac{d^n}{dx^n}(x^2-1)^n\,dx$$
you can integral by part n time
So..
$$ u= \frac{d^n}{dx^n}(x^2-1)^n$$ and $$du=\frac{d^{n+1}}{dx^{n+1}}(x^2-1)^n\,dx$$
$$ dv= \frac{d^n}{dx^n}(x^2-1)^ndx$$ and $$v=\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n$$
$$ udv=uv-vdu$$
$$ uv= \frac{d^n}{dx^n}(x^2-1)^n \frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n =0 $$ When limit of the integrate from -1 to 1
So when integrate by part n times
$$ \int_{-1}^{1} P_{n}(x)P_{n}(x)\,dx= \frac{1}{2^{2n}(n!)^2}\left[0+0+...+(-1)^n\int_{-1}^{1}\frac{d^{n-n}}{dx^{n-n}}(x^2-1)^n \frac{d^{n+n}}{dx^{n+n}}(x^2-1)^n \,dx\right]$$
$$ \int_{-1}^{1} P_{n}(x)P_{n}(x)\,dx= \frac{1}{2^{2n}(n!)^2}\left[\int_{-1}^{1}(x^2-1)^n \frac{d^{2n}}{dx^{2n}}(1-x^2)^n \,dx\right]$$ ; times (-1)^n in (x^n-1)^n
You know $$(1-x^2)^n=\sum_{k=0}^{n}{n \choose k} 1^{n-k}(-1)^n(x^2)^n $$
$$ \int_{-1}^{1} P_{n}(x)P_{n}(x)\,dx= \frac{1}{2^{2n}(n!)^2}\left[\int_{-1}^{1}(x^2-1)^n \frac{d^{2n}}{dx^{2n}}\sum_{k=0}^{n}{n \choose k} 1^{n-k}(-1)^n(x^2)^n \,dx\right]$$
The final term of $$(1-x^2)^n=\sum_{k=0}^{n}{n \choose k} 1^{n-k}(-1)^n(x^2)^n = ...+ {n \choose n }(-1)^n x^{2n}$$
because $$ \frac{d^m}{dx^m}x^n = 0$$ when n < m
So that
$$ \int_{-1}^{1} P_{n}(x)P_{n}(x)\,dx= \frac{1}{2^{2n}(n!)^2}\left[\int_{-1}^{1}(x^2-1)^n \frac{d^{2n}}{dx^{2n}}(-1)^nx^{2n} \,dx\right]$$
$$ \int_{-1}^{1} P_{n}(x)P_{n}(x)\,dx= \frac{1}{2^{2n}(n!)^2}\left[\int_{-1}^{1}(x^2-1)^n (-1)^n (2n)! \,dx\right]$$
$$ \int_{-1}^{1} P_{n}(x)P_{n}(x)\,dx= \frac{(2n)!}{2^{2n}(n!)^2}\left[\int_{-1}^{1}(1-x^2)^n \,dx\right]$$
set $$s =\frac{x+1}{2}$$
$$ \int_{-1}^{1} P_{n}(x)P_{n}(x)\,dx= \frac{(2n)!}{2^{2n}(n!)^2}\left[\int_{0}^{1} 2\cdot 2^{2n}s^n(1-s)^n\,ds\right]$$ beta function
$$ \int_{-1}^{1} P_{n}(x)P_{n}(x)\,dx= \frac{(2n)!}{2^{2n}(n!)^2} 2\cdot 2^{2n}\frac{(n!)^2}{(2n+1)!}$$
$$ \int_{-1}^{1} P_{n}(x)P_{n}(x)\,dx= \frac{2}{2n+1}$$
see that
$$ \int_{-1}^{1} cos(x)P_{n}(x)\,dx= \frac{1}{2^{n}n!}\int_{-1}^{1}cos(x)\frac{d^n}{dx^n}(x^2-1)^n\,dx$$
You can integral by part n times.
$$ u = con(x)$$ and $$ dv = \frac{d^n}{dx^n}(x^2-1)^n dx $$
So . . .
$$ \int_{-1}^{1} cos(x)P_{n}(x)\,dx= \frac{(-1)^n}{2^{n}n!}\int_{-1}^{1}(x^2-1)^n \frac{d^n}{dx^n}cos(x),dx$$
give two solution
in case n = odd number
$$\int_{-1}^{1} cos(x)P_{n}(x)\,dx=\frac{1}{2^nn!}(-1)^n\int_{-1}^{1} (-1)^{\frac{n+1}{2}}sin(x) (x^2-1)^n\,dx$$
in case n = even number
$$\int_{-1}^{1} cos(x)P_{n}(x)\,dx=\frac{1}{2^nn!}(-1)^n\int_{-1}^{1} (-1)^{\frac{n}{2}}cos(x) (x^2-1)^n\,dx$$
But $$ sin(x) (x^2-1)^n $$ are Odd function So $$ \frac{1}{2^nn!}(-1)^n\int_{-1}^{1} (-1)^{\frac{n+1}{2}}sin(x) (x^2-1)^n\,dx = 0$$
left only one solution
$$\int_{-1}^{1} cos(x)P_{n}(x)\,dx=\frac{1}{2^nn!}(-1)^n\int_{-1}^{1} (-1)^{\frac{n}{2}}cos(x) (x^2-1)^n\,dx$$ n = 0,2,4,...
$$\int_{-1}^{1} cos(x)P_{n}(x)\,dx=\frac{1}{2^nn!}(-1)^{\frac{n}{2}} 2 \int_{0}^{1} cos(x) (1-x^2)^n\,dx$$ ; even function
Wolfram|Alpha: Computational Intelligence
You know $$ cos(x) = \sum_{i=0}^{\infty}\frac{(-1)^i}{(2i)!}x^{2i} $$
$$\int_{-1}^{1} cos(x)P_{n}(x)\,dx=\frac{2}{2^nn!}(-1)^{\frac{n}{2}} \int_{0}^{1} \sum_{i=0}^{\infty}\frac{(-1)^i}{(2i)!}x^{2i} (1-x^2)^n\,dx$$
$$\int_{-1}^{1} cos(x)P_{n}(x)\,dx=\frac{1}{2^nn!}(-1)^{\frac{n}{2}} \sum_{i=0}^{\infty}\frac{(-1)^i}{(2i)!} (2)\int_{0}^{1} x^{2i} (1-x^2)^n\,dx$$; Beta function form
$$\int_{-1}^{1} cos(x)P_{n}(x)\,dx=\frac{1}{2^nn!}(-1)^{\frac{n}{2}} \sum_{i=0}^{\infty}\frac{(-1)^i}{(2i)!} \frac{(i-\frac{1}{2})!n!}{(i-\frac{1}{2}+n+1)!}$$
$$\int_{-1}^{1} cos(x)P_{n}(x)\,dx=\frac{1}{2^nn!}(-1)^{\frac{n}{2}} \sum_{i=0}^{\infty}\frac{(-1)^i}{(2i)!} \frac{(i-\frac{1}{2})! \Gamma(n+1)}{(i+n+\frac{1}{2})!}$$
from $$(i-\frac{1}{2})! = \frac{1}{i!}(\frac{1}{2})^{2i}\sqrt{\pi}(2i)!$$
$$\int_{-1}^{1} cos(x)P_{n}(x)\,dx=\frac{1}{2^nn!}(-1)^{\frac{n}{2}}\Gamma(n+1) \sum_{i=0}^{\infty}\frac{(-1)^i}{(2i)!} \frac{1}{(i+n+\frac{1}{2})!}\frac{1}{i!}(\frac{1}{2})^{2i}\sqrt{\pi}(2i)!$$
$$\int_{-1}^{1} cos(x)P_{n}(x)\,dx=\frac{1}{2^nn!}(-1)^{\frac{n}{2}}\Gamma(n+1) \sum_{i=0}^{\infty}\frac{(-1)^i}{(2i)!} \frac{1}{(i+n+\frac{1}{2})!}\frac{1}{i!}(\frac{1}{2})^{2i}\sqrt{\pi}(2i)! \cdot \frac{2^{n+\frac{1}{2}}}{2^{n+\frac{1}{2}}}$$
$$\int_{-1}^{1} cos(x)P_{n}(x)\,dx=\frac{1}{2^nn!}(-1)^{\frac{n}{2}}\Gamma(n+1) 2^{n+\frac{1}{2}}\sqrt{\pi}\sum_{i=0}^{\infty}\frac{(-1)^i}{i!} \frac{1}{(i+n+\frac{1}{2})!}(\frac{1}{2})^{2i} \cdot (\frac{1}{2})^{n+\frac{1}{2}}$$
$$\int_{-1}^{1} cos(x)P_{n}(x)\,dx=\frac{1}{2^nn!}(-1)^{\frac{n}{2}}\Gamma(n+1) 2^{n+\frac{1}{2}}\sqrt{\pi}\sum_{i=0}^{\infty}\frac{(-1)^i}{i!} \frac{1}{(i+n+\frac{1}{2})!}(\frac{1}{2})^{n+\frac{1}{2}+2i}$$
$$\int_{-1}^{1} cos(x)P_{n}(x)\,dx=\frac{1}{2^nn!}(-1)^{\frac{n}{2}}\Gamma(n+1) 2^{n+\frac{1}{2}}\sqrt{\pi}J_{n+\frac{1}{2}}(1)$$ ; Bessel polynomialSo
$$C_{n}=\frac{2n+1}{2}\int_{-1}^{1} cos(x)P_{n}(x)\,dx= \frac{2n+1}{2} \frac{1}{2^nn!}(-1)^{\frac{n}{2}}\Gamma(n+1) 2^{n+\frac{1}{2}}\sqrt{\pi}J_{n+\frac{1}{2}}(1)$$
$$C_{n}=\frac{2n+1}{2}\int_{-1}^{1} cos(x)P_{n}(x)\,dx= \frac{2n+1}{n!}(-1)^{\frac{n}{2}}\Gamma(n+1) 2^{-\frac{1}{2}}\sqrt{\pi}J_{n+\frac{1}{2}}(1)$$ n = 0,2,4,...
and finaly
$$ cos(x) = \sum_{n=0}^{\infty}C_{n}P_{n}(x)$$
$$ cos(x) = \sum_{n=0}^{\infty} \frac{2n+1}{n!}(-1)^{\frac{n}{2}}\Gamma(n+1) 2^{-\frac{1}{2}}\sqrt{\pi}J_{n+\frac{1}{2}}(1) P_{n}(x)$$
- My solution