Question

If $u_n={\int }_0^{\pi /2}\frac{1-cos2nx}{1-cos2x}dx$, then find the value of the determinant $\Delta =\left|\begin{array}{ccc}{u}_1& u_2& u_3\\ u_4& u_5& u_6\\ u_7& u_8& u_9\end{array}\right|$

Answer 1

$u_n={\int }_0^{\frac{\pi }{2}}\frac{1-cos2nx}{1-cos2x}dx{u}_n+u_{n+2}-2u_{n+1}={\int }_0^{\frac{\pi }{2}}\frac{(1-cos2nx)+(1-cos(2n+4\left)x\right)-2(1-cos(2n+2\left)x\right)}{1-cos2x}dx={\int }_0^{\frac{\pi }{2}}\frac{-2\cos (2n+2)x\cos 2x+2\cos (2n+2)x}{1-\cos 2x}dx={\int }_0^{\frac{\pi }{2}}2\cos (2n+2)xdx=2{\left[\frac{\sin (2n+2)x}{2n+2}\right]}_0^{\frac{\pi }{2}}=0\therefore u_n+u_{n+2}-2u_{n+1}=0\Delta =\left|\begin{array}{ccc}{u}_1& u_2& u_3\\ u_4& u_5& u_6\\ u_7& u_8& u_9\end{array}\right|applying{C}_1\to C_1+C_3-2C_2\Delta =\left|\begin{array}{ccc}{u}_1+u_3-2u_2& u_2& u_3\\ u_4+u_6-2u_5& u_5& u_6\\ u_7+u_9-2u_8& u_8& u_9\end{array}\right|=\left|\begin{array}{ccc}0& u_2& u_3\\ 0& u_5& u_6\\ 0& u_8& u_9\end{array}\right|=0$