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Question Number 184988 by saboorhalimi last updated on 15/Jan/23

Answered by a.lgnaoui last updated on 15/Jan/23

$$\mathrm{\Omega }={\int }_0^{\frac{\pi }{2}}(k\cos \left(nx\right)+m\sin nx)dx+$$$${\int }_{\frac{\pi }{2}}^{\pi }(k\cos \left(nx\right)+m\sin nx)dx$$$$={[\frac{k}{n}\sin nx-\frac{m}{n}\cos nx]}_0^{\frac{\pi }{2}}+$$$${[\frac{k}{n}\sin nx-\frac{m}{n}\cos nx]}_{\frac{\pi }{2}}^{\pi }=$$$$\frac{1}{\mathit{n}}{\left[(k\sin \frac{n\pi }{2}-m\cos \frac{\pi n}{2})\right]}_0^{\frac{\pi }{2}}+$$$$\frac{1}{n}{\left[(k\sin \frac{\pi n}{2}-m\cos \frac{\pi n}{2})\right]}_{\frac{\pi }{2}}^{\pi }$$$$$$$$p>0npair\mathrm{\Omega }=0$$$$\mathrm{\Omega }=\frac{m}{n}\sqrt{2}$$$$p⩾1\left(nimpair\right)$$$$\mathrm{\Omega }=\frac{k}{n}+\frac{k}{n}{\left[\sin \left(nx\right)\right]}_{\frac{\pi }{2}}^{\pi }=\frac{k}{n}-\frac{k}{n}=0$$$$$$

Commented by Frix last updated on 15/Jan/23

$$\mathrm{Error}:\mathrm{\Omega }\ne \underset{0}{\stackrel{\frac{\pi }{2}}{\int }}...-\underset{\frac{\pi }{2}}{\stackrel{\pi }{\int }}...\mathrm{because}\mathrm{the}\mathrm{zeros}\mathrm{of}$$$$k\cos nx+m\sin nx\mathrm{are}x=\frac{z\pi -{\tan }^{-1}\frac{k}{m}}{n}\mathrm{\forall }z\in \mathbb{Z}$$$$\mathrm{and}\mathrm{not}\mathrm{at}x=\frac{\pi }{2}.$$

Commented by saboorhalimi last updated on 16/Jan/23

$$\mathbf{please}\mathbf{check}\mathbf{your}\mathbf{solution}\mathbf{again}$$

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