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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
With differential equation.
$${With}\:{differential}\:{equation}. \\ $$
Answered by Kamel last updated on 16/Jan/22
Commented by Kamel last updated on 16/Jan/22
With Laplas transform′s
$${With}\:{Laplas}\:{transform}'{s} \\ $$
Answered by mindispower last updated on 18/Jan/22
sin(x)=−ish(ix) =−i∫_0 ^π cot((x/2))sh(iasin(x))sh(acos(x))dx =−i∫_0 ^π cot((x/2))(ch(ae^(ix) )−ch(ae^(−ix) ) =2Im∫_0 ^π cot((x/2))ch(ae^(ix) ) =2Im∫_0 ^π cot((x/2))Σ_(n≥0) (((ae^(ix) )^(2n) )/((2n)!)) =2Σ_(n≥1) a^(2n) ∫_0 ^π ((sin(2nx))/((2n)!))cot((x/2))dx =Σ_(n≥1) (a^(2n) /((2n)!))∫_0 ^π sin(2nx)(((cos((x/2)))/(sin((x/2))))−((sin((x/2)))/(cos((x/2))))) =2Σ_(n≥1) (a^(2n) /((2n)!))∫_0 ^π ((sin(2nx)cos(x))/(sin(x)))dx=Ω ∫_0 ^π ((sin(2nx)cos(x))/(sin(x)))dx=∫_π ^(2π) ((sin(2nx)cos(x))/(sin(x)))dx Ω=Σ_(n≥1) (a^(2n) /((2n)!))∫_0 ^(2π) ((sin(2nx)cos(x))/(sin(x))) I_(2n+2) −I_(2n) =∫_0 ^(2π) ((sin(2n+2x)−sin(2nx))/(sin(x)))cos(x) sin(2n+2)x−sin(2nx)=2sin(x)coos(n+1)x =∫_0 ^(2π) 2cos(n+1)xcos(x) =∫_0 ^(2π) cos(nx)−cos(n+2)x dx=0 ∀n∈N^∗ I_(2n+2) =I_(2n) ⇔I_(2n) =I_2 ⇔I_(2n) =∫_0 ^(2π) ((sin(2nx)cos(x))/(sin(x)))dx=∫_0 ^(2π) ((sin(2x)cos(x))/(sin(x)))dx ∫_0 ^(2π) 2cos^2 (x)dx=∫_0 ^(2π) (1+cos(2x))dx=2π we get Σ_(n≥1) (a^(2n) /((2n)!)).π =π(Σ_(n≥0) (a^(2n) /((2n)!))−1)=π(ch(a)−1)
$${sin}\left({x}\right)=−{ish}\left({ix}\right) \\ $$$$=−{i}\int_{\mathrm{0}} ^{\pi} {cot}\left(\frac{{x}}{\mathrm{2}}\right){sh}\left({iasin}\left({x}\right)\right){sh}\left({acos}\left({x}\right)\right){dx} \\ $$$$=−{i}\int_{\mathrm{0}} ^{\pi} {cot}\left(\frac{{x}}{\mathrm{2}}\right)\left({ch}\left({ae}^{{ix}} \right)−{ch}\left({ae}^{−{ix}} \right)\right. \\ $$$$=\mathrm{2}{Im}\int_{\mathrm{0}} ^{\pi} {cot}\left(\frac{{x}}{\mathrm{2}}\right){ch}\left({ae}^{{ix}} \right) \\ $$$$=\mathrm{2}{Im}\int_{\mathrm{0}} ^{\pi} {cot}\left(\frac{{x}}{\mathrm{2}}\right)\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\left({ae}^{{ix}} \right)^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!} \\ $$$$=\mathrm{2}\underset{{n}\geqslant\mathrm{1}} {\sum}{a}^{\mathrm{2}{n}} \int_{\mathrm{0}} ^{\pi} \frac{{sin}\left(\mathrm{2}{nx}\right)}{\left(\mathrm{2}{n}\right)!}{cot}\left(\frac{{x}}{\mathrm{2}}\right){dx} \\ $$$$=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{a}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}\int_{\mathrm{0}} ^{\pi} {sin}\left(\mathrm{2}{nx}\right)\left(\frac{{cos}\left(\frac{{x}}{\mathrm{2}}\right)}{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}−\frac{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}{{cos}\left(\frac{{x}}{\mathrm{2}}\right)}\right) \\ $$$$=\mathrm{2}\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{a}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}\int_{\mathrm{0}} ^{\pi} \frac{{sin}\left(\mathrm{2}{nx}\right){cos}\left({x}\right)}{{sin}\left({x}\right)}{dx}=\Omega \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{sin}\left(\mathrm{2}{nx}\right){cos}\left({x}\right)}{{sin}\left({x}\right)}{dx}=\int_{\pi} ^{\mathrm{2}\pi} \frac{{sin}\left(\mathrm{2}{nx}\right){cos}\left({x}\right)}{{sin}\left({x}\right)}{dx} \\ $$$$\Omega=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{a}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{sin}\left(\mathrm{2}{nx}\right){cos}\left({x}\right)}{{sin}\left({x}\right)} \\ $$$${I}_{\mathrm{2}{n}+\mathrm{2}} −{I}_{\mathrm{2}{n}} =\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{sin}\left(\mathrm{2}{n}+\mathrm{2}{x}\right)−{sin}\left(\mathrm{2}{nx}\right)}{{sin}\left({x}\right)}{cos}\left({x}\right) \\ $$$${sin}\left(\mathrm{2}{n}+\mathrm{2}\right){x}−{sin}\left(\mathrm{2}{nx}\right)=\mathrm{2}{sin}\left({x}\right){coos}\left({n}+\mathrm{1}\right){x} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{2}{cos}\left({n}+\mathrm{1}\right){xcos}\left({x}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{2}\pi} {cos}\left({nx}\right)−{cos}\left({n}+\mathrm{2}\right){x}\:{dx}=\mathrm{0} \\ $$$$\forall{n}\in\mathbb{N}^{\ast} \:{I}_{\mathrm{2}{n}+\mathrm{2}} ={I}_{\mathrm{2}{n}} \Leftrightarrow{I}_{\mathrm{2}{n}} ={I}_{\mathrm{2}} \\ $$$$\Leftrightarrow{I}_{\mathrm{2}{n}} =\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{sin}\left(\mathrm{2}{nx}\right){cos}\left({x}\right)}{{sin}\left({x}\right)}{dx}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{sin}\left(\mathrm{2}{x}\right){cos}\left({x}\right)}{{sin}\left({x}\right)}{dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{2}{cos}^{\mathrm{2}} \left({x}\right){dx}=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \left(\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)\right){dx}=\mathrm{2}\pi \\ $$$${we}\:{get}\:\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{a}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}.\pi \\ $$$$=\pi\left(\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{{a}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}−\mathrm{1}\right)=\pi\left({ch}\left({a}\right)−\mathrm{1}\right) \\ $$$$ \\ $$
Commented by HongKing last updated on 20/Jan/22
perfect my dear Sir thank you
$$\mathrm{perfect}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you} \\ $$

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