Parabolic sine approximation Problem Find a parabola ($f(x)=ax^2+bx+c$) that approximate the function sine the best on interval [0,$\pi$].
The distance between two solutions is calculated this way (in relation to scalar product): $\langle u,v \rangle=\int_0^\pi fg$. 
My (wrong) solution I thought that I would get the solution by calculating the orthogonal projection $w=a+bx+cx^2$ of $v=\sin x$ on subspace $W=\langle u_1,u_2,u_3 \rangle=\langle 1,x,x^2\rangle$ using Gramm matrix. Then I have
$$\begin{pmatrix}\langle u_1, u_1\rangle&\langle u_1, u_2\rangle&\langle u_1, u_3\rangle\\
\langle u_2, u_1\rangle&\langle u_2, u_2\rangle&\langle u_2, u_3\rangle\\
\langle u_3, u_1\rangle&\langle u_3, u_2\rangle& \langle u_3, u_3\rangle\end{pmatrix}
\begin{pmatrix} a\\b\\c\end{pmatrix}=
\begin{pmatrix} \langle u_1, v\rangle\\
\langle u_2,v\rangle\\
\langle u_3, v\rangle \end{pmatrix}$$
So then
$$\begin{pmatrix} \int_0^\pi 1&\int_0^\pi x&\int_0^\pi x^2\\
\int_0^\pi x& \int_0^\pi x^2& \int_0^\pi x^3 \\
\int_0^\pi x^2&\int_0^\pi x^3&\int_0^\pi x^4\end{pmatrix}\begin{pmatrix} a\\b\\c\end{pmatrix}=\begin{pmatrix} \int_0^\pi \sin x\\\int_0^\pi x\sin x\\\int_0^\pi x^2 \sin x \end{pmatrix}$$
But solving these equations didn't give me any good answer. So my question is - is my way of solving it completely wrong (if so, can you give me hints how to do it otherwise)?
Thank you
Edit
Then$$ 
\begin{pmatrix} \pi&\frac{\pi^2}{2}&\frac{\pi^3}{3}\\
\frac{\pi^2}{2}&\frac{\pi^3}{3}&\frac{\pi^4}{4}\\
\frac{\pi^3}{3}&\frac{\pi^4}{4}&\frac{\pi^5}{5}\end{pmatrix}\begin{pmatrix} a\\b\\c\end{pmatrix}=\begin{pmatrix} 2\\\pi\\\pi^2-4 \end{pmatrix}$$
$$ 
\begin{pmatrix} \pi&\frac{\pi^2}{2}&\frac{\pi^3}{3}\\
0&\frac{\pi^3}{12}&\frac{\pi^4}{12}\\
0&\frac{\pi^4}{12}&\frac{4\pi^5}{45}\end{pmatrix}\begin{pmatrix} a\\b\\c\end{pmatrix}=\begin{pmatrix} 2\\0\\\frac{1}{3}\pi^2-4 \end{pmatrix}$$
$$ 
\begin{pmatrix} \pi&\frac{\pi^2}{2}&\frac{\pi^3}{3}\\
0&\frac{\pi^3}{12}&\frac{\pi^4}{12}\\
0&0&\frac{\pi^5}{180}\end{pmatrix}\begin{pmatrix} a\\b\\c\end{pmatrix}=\begin{pmatrix} 2\\0\\\frac{1}{3}\pi^2-4 \end{pmatrix}$$
Edit II My way (I have found the mistake I did) will get the result
$$f(x)=\dfrac{60(\pi^2-12)}{\pi^5}x^2-\dfrac{60(\pi^2-12)}{\pi^4}x+\dfrac{12(\pi^2-10)}{\pi^3}$$
which I hope is the right answer with error approx. $0,000936$

First,I donnot think the orthogonal projection is proper here. The orthogonal projections are often used to make infinity series to approach function,because the coefficients there is more simple,you can try to use error square or the maximun of error
