Question Number 140282 by qaz last updated on 06/May/21
$$\underset{{k}=\mathrm{0}} {\overset{{p}−\mathrm{1}} {\sum}}\begin{pmatrix}{{p}}\\{{k}}\end{pmatrix}\mathrm{sin}\:\left[\mathrm{2}\left({p}−{k}\right){x}\right]=? \\ $$$$\begin{pmatrix}{{p}}\\{\mathrm{0}}\end{pmatrix}\mathrm{sin}\:\left(\mathrm{2}{px}\right)+\begin{pmatrix}{{p}}\\{\mathrm{1}}\end{pmatrix}\mathrm{sin}\:\left[\left(\mathrm{2}{p}−\mathrm{2}\right){x}\right]+\begin{pmatrix}{{p}}\\{\mathrm{2}}\end{pmatrix}\mathrm{sin}\:\left[\left(\mathrm{2}{p}−\mathrm{4}\right){x}\right]+…+\begin{pmatrix}{\:\:\:{p}}\\{{p}−\mathrm{1}}\end{pmatrix}\mathrm{sin}\:\left(\mathrm{2}{x}\right)=\mathrm{2}^{{p}} \centerdot\mathrm{cos}\:^{{p}} \left({x}\right)\centerdot\mathrm{sin}\:\left({px}\right)\:\:\:\:\:??? \\ $$$${or}\:\:\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{cos}\:^{{p}} \left({x}\right)\centerdot\mathrm{sin}\:\left({px}\right)}{{x}}{dx}=\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\mathrm{2}^{−{p}} \right)\:\:\:\:\:{why}\:??? \\ $$