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Question Number 164418 by HongKing last updated on 16/Jan/22

Commented by Kamel last updated on 16/Jan/22

Commented by Kamel last updated on 16/Jan/22

$$Withdifferentialequation.$$

Answered by Kamel last updated on 16/Jan/22

Commented by Kamel last updated on 16/Jan/22

$$WithLaplas{transform}^{\prime }s$$

Answered by mindispower last updated on 18/Jan/22

$$sin\left(x\right)=-ish\left(ix\right)$$$$=-i{\int }_0^{\pi }cot\left(\frac{x}{2}\right)sh\left(iasin\left(x\right)\right)sh\left(acos\left(x\right)\right)dx$$$$=-i{\int }_0^{\pi }cot\left(\frac{x}{2}\right)(ch\left({ae}^{ix}\right)-ch\left({ae}^{-ix}\right)$$$$=2Im{\int }_0^{\pi }cot\left(\frac{x}{2}\right)ch\left({ae}^{ix}\right)$$$$=2Im{\int }_0^{\pi }cot\left(\frac{x}{2}\right)\sum _{n⩾0}\frac{{\left({ae}^{ix}\right)}^{2n}}{\left(2n\right)!}$$$$=2\sum _{n⩾1}{a}^{2n}{\int }_0^{\pi }\frac{sin\left(2nx\right)}{\left(2n\right)!}cot\left(\frac{x}{2}\right)dx$$$$=\sum _{n⩾1}\frac{a^{2n}}{\left(2n\right)!}{\int }_0^{\pi }sin\left(2nx\right)(\frac{cos\left(\frac{x}{2}\right)}{sin\left(\frac{x}{2}\right)}-\frac{sin\left(\frac{x}{2}\right)}{cos\left(\frac{x}{2}\right)})$$$$=2\sum _{n⩾1}\frac{a^{2n}}{\left(2n\right)!}{\int }_0^{\pi }\frac{sin\left(2nx\right)cos\left(x\right)}{sin\left(x\right)}dx=\mathrm{\Omega }$$$$$$$${\int }_0^{\pi }\frac{sin\left(2nx\right)cos\left(x\right)}{sin\left(x\right)}dx={\int }_{\pi }^{2\pi }\frac{sin\left(2nx\right)cos\left(x\right)}{sin\left(x\right)}dx$$$$\mathrm{\Omega }=\sum _{n⩾1}\frac{a^{2n}}{\left(2n\right)!}{\int }_0^{2\pi }\frac{sin\left(2nx\right)cos\left(x\right)}{sin\left(x\right)}$$$$I_{2n+2}-I_{2n}={\int }_0^{2\pi }\frac{sin(2n+2x)-sin\left(2nx\right)}{sin\left(x\right)}cos\left(x\right)$$$$sin(2n+2)x-sin\left(2nx\right)=2sin\left(x\right)coos(n+1)x$$$$={\int }_0^{2\pi }2cos(n+1)xcos\left(x\right)$$$$={\int }_0^{2\pi }cos\left(nx\right)-cos(n+2)xdx=0$$$$\mathrm{\forall }n\in {\mathbb{N}}^{\ast }{I}_{2n+2}=I_{2n}\iff I_{2n}=I_2$$$$\iff I_{2n}={\int }_0^{2\pi }\frac{sin\left(2nx\right)cos\left(x\right)}{sin\left(x\right)}dx={\int }_0^{2\pi }\frac{sin\left(2x\right)cos\left(x\right)}{sin\left(x\right)}dx$$$${\int }_0^{2\pi }2{cos}^2\left(x\right)dx={\int }_0^{2\pi }(1+cos\left(2x\right))dx=2\pi$$$$weget\sum _{n⩾1}\frac{a^{2n}}{\left(2n\right)!}.\pi$$$$=\pi (\sum _{n⩾0}\frac{a^{2n}}{\left(2n\right)!}-1)=\pi (ch\left(a\right)-1)$$$$$$

Commented by HongKing last updated on 20/Jan/22

$$\mathrm{perfect}\mathrm{my}\mathrm{dear}\mathrm{Sir}\mathrm{thank}\mathrm{you}$$

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