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Question Number 94914 by 174 last updated on 21/May/20

Answered by mathmax by abdo last updated on 22/May/20

$${\mathrm{A}}_{+}=\int {\mathrm{x}}^2{({\mathrm{a}}^2+{\mathrm{x}}^2)}^{\mathrm{m}-1}\mathrm{dx}\left(\mathrm{m}\mathrm{integr}\right)\Rightarrow \mathrm{x}=\mathrm{at}$$$${\mathrm{A}}_{+}=\int {\mathrm{a}}^2{\mathrm{t}}^2{\mathrm{a}}^{2\mathrm{m}-2}{(1+{\mathrm{t}}^2)}^{\mathrm{m}-1}\mathrm{adt}$$$$={\mathrm{a}}^{2\mathrm{m}+1}\int {\mathrm{t}}^2\sum _{\mathrm{k}=0}^{\mathrm{m}-1}{\mathrm{C}}_{\mathrm{m}-1}^{\mathrm{k}}{\mathrm{t}}^{2\mathrm{k}}\mathrm{dt}$$$$={\mathrm{a}}^{2\mathrm{m}+1}\sum _{\mathrm{k}=0}^{\mathrm{m}-1}{\mathrm{C}}_{\mathrm{m}-1}^{\mathrm{k}}{\mathrm{t}}^{2\mathrm{k}+2}\mathrm{dt}$$$$={\mathrm{a}}^{2\mathrm{m}+1}\sum _{\mathrm{k}=0}^{\mathrm{m}-1}\frac{{\mathrm{C}}_{\mathrm{m}-1}^{\mathrm{k}}}{2\mathrm{k}+3}{\mathrm{t}}^{2\mathrm{k}+3}+\mathrm{C}$$$$={\mathrm{a}}^{3\mathrm{m}+1}\sum _{\mathrm{k}=0}^{\mathrm{m}-1}\frac{{\mathrm{C}}_{\mathrm{m}+1}^{\mathrm{k}}}{2\mathrm{k}+3}{\left(\frac{\mathrm{x}}{\mathrm{a}}\right)}^{3\mathrm{k}+3}+\mathrm{C}(\mathrm{we}\mathrm{suppose}\mathrm{a}\ne 0)$$$${\mathrm{A}}_{-}=\int {\mathrm{x}}^2{({\mathrm{a}}^2-{\mathrm{x}}^2)}^{\mathrm{m}-1}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}=\mathrm{asin}\alpha$$$$\Rightarrow {\mathrm{A}}_{-}=\int {\mathrm{a}}^2{\sin }^2\alpha {\mathrm{a}}^{2\mathrm{m}-2}{\left({\cos }^2\alpha \right)}^{\mathrm{m}-1}\mathrm{a}\cos \alpha \mathrm{d}\alpha$$$$={\mathrm{a}}^{2\mathrm{m}+1}\int {\sin }^2\alpha {\cos }^{2\mathrm{m}-1}\alpha \mathrm{d}\alpha$$$$={\mathrm{a}}^{2\mathrm{m}+1}\int (1-{\cos }^2\alpha ){\cos }^{2\mathrm{m}-1}\alpha \mathrm{d}\alpha$$$$={\mathrm{a}}^{2\mathrm{m}+1}\int {\cos }^{2\mathrm{m}-1}\alpha \mathrm{d}\alpha -{\mathrm{a}}^{2\mathrm{m}+1}\int {\cos }^{2\mathrm{m}+1}\alpha \mathrm{d}\alpha \left(\mathrm{wallis}\mathrm{integral}\right)$$$$\int {\cos }^{2\mathrm{m}-1}\alpha \mathrm{d}\alpha =\int {\left(\frac{{\mathrm{e}}^{\mathrm{i}\alpha }+{\mathrm{e}}^{-\mathrm{i}\alpha }}{2}\right)}^{2\mathrm{m}-1}\mathrm{d}\alpha$$$$=\frac{1}{2^{2\mathrm{m}-1}}\int \sum _{\mathrm{k}=0}^{2\mathrm{m}-1}{\mathrm{C}}_{2\mathrm{m}-1}^{\mathrm{k}}{\mathrm{e}}^{\mathrm{ik}\alpha }\times {\mathrm{e}}^{-\mathrm{i}(2\mathrm{m}-1-\mathrm{k})\alpha }\mathrm{d}\alpha$$$$=\frac{1}{2^{2\mathrm{m}-1}}\sum _{\mathrm{k}=0}^{2\mathrm{m}-1}{\mathrm{C}}_{2\mathrm{m}-1}^{\mathrm{k}}\int {\mathrm{e}}^{\mathrm{i}(\mathrm{k}-2\mathrm{m}+1+\mathrm{k})\alpha }\mathrm{d}\alpha$$$$=\frac{1}{2^{2\mathrm{m}-1}}\sum _{\mathrm{k}=0}^{2\mathrm{m}-1}{\mathrm{C}}_{2\mathrm{m}-1}^{\mathrm{k}}\times \frac{1}{2\mathrm{k}-2\mathrm{m}+1}{\mathrm{e}}^{\mathrm{i}(2\mathrm{k}-2\mathrm{m}+1)\alpha }+\mathrm{C}$$$$$$$$$$

Commented by 174 last updated on 22/May/20

thanks a lot

Commented by ElOuafi last updated on 22/May/20

$$perfectsir!!thankyou.$$

Commented by mathmax by abdo last updated on 22/May/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}$$

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