Finding Chebychev coefficients

Question

To find Chebyshev coefficients I need to compute the following polynomial with $T_k(x)$ being the Chebyshev polynomial. We use that $T_k(\cos(z))=\cos(kz)$ $$ 2\int_0^1 T_k(x)\frac{1}{\sqrt{1-x^2}} dx= $$ Now we change the variable $x=\cos(z)\Leftrightarrow z=\cos^{-1}(x)$ and $dz=\cos^{-1}(x)$, $dx=-\frac{1}{\sqrt{1-x^2}}dx$. Wouldn't we have because $\cos^{-1}(1)=0,\cos^{-1}(0)=\pi /2$ $$ 2 \int_0^{\pi/2}-\cos(kz) dz $$ I am wondering because my solution says $$ 2 \int_0^{\pi/2}\cos(kz) dz $$ so I am confused where the "-" is going and would appreciate your help :)

Answer 1

When you apply the substitution formula, the lower bounds and the upper bounds must correspond, so you obtain $$2\int_0^1 T_k(x)\frac{1}{\sqrt{1-x^2}}\,\mathrm dx=2 \int^0_{\pi/2}\mkern-9mu-\cos(kz)\,\mathrm dz=2 \int_0^{\pi/2}\mkern -9mu\cos(kz) \,\mathrm dz$$