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Question Number 185058 by saboorhalimi last updated on 16/Jan/23

Answered by Frix last updated on 16/Jan/23

$$k\cos nx+m\sin nx=\sqrt{k^2+m^2}\sin (nx+{\tan }^{-1}\frac{k}{m})$$$$\mathrm{\Omega }=\sqrt{k^2+m^2}\underset{0}{\stackrel{\pi }{\int }}\mid \sin (nx+{\tan }^{-1}\frac{k}{m})\mid dx=$$$$=\frac{\sqrt{k^2+m^2}}{n}\times \underset{{\tan }^{-1}\frac{k}{m}}{\stackrel{n\pi +{\tan }^{-1}\frac{k}{m}}{\int }}\mid \sin t\mid dt=$$$$=\frac{\sqrt{k^2+m^2}}{n}\times \underset{0}{\stackrel{n\pi }{\int }}\mid \sin t\mid dt=\sqrt{k^2+m^2}\underset{0}{\stackrel{\pi }{\int }}\sin tdt=$$$$=2\sqrt{k^2+m^2}$$

Commented by saboorhalimi last updated on 16/Jan/23

$$\mathit{n}\mathit{i}\mathit{c}\mathit{e}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}$$

Commented by Frix last updated on 16/Jan/23

$$\mathrm{Thank}\mathrm{you},\mathrm{I}\mathrm{added}\mathrm{some}\mathrm{lines}.$$

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