Find the cubic near-minimax or Chebyshev approximation for $f(x) = \sin(x)$ on the interval $[0,\frac{\pi}{2}]$.
Attempt: The first four Chebyshev polynomials are
\begin{align} T_0(x)&=1,\\ T_1(x)&=x,\\ T_2(x)&=2x^2-1,\\ T_3(x)&=4x^3-x. \end{align}
Then $\sin(x) \approx c_0 T_0 +c_1 T_1 +c_2 T_2 +c_3 T_3$, where $$c_0= \frac{1}{\pi} \int_{0}^{\pi/2} \frac{1}{\sqrt{1-x^2}}dx .$$
But $\displaystyle c_1= \frac{2}{\pi} \int_{0}^{\pi/2} \frac{x}{\sqrt{1-x^2}}dx$ is complex.
So how can I find the constants? Am I doing it right? Any help to solve this problem will be appreciated.