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Question Number 127885 by psyche last updated on 02/Jan/21
∫(((sin (2tan^(−1) (x)+x))/x)) the limit [0,∞)
$$\int\left(\frac{\mathrm{sin}\:\left(\mathrm{2tan}^{−\mathrm{1}} \left({x}\right)+{x}\right)}{{x}}\right)\:\:{the}\:{limit}\:\left[\mathrm{0},\infty\right) \\ $$
Answered by Lordose last updated on 02/Jan/21
Ω = ∫_0 ^( ∞) ((sin(2tan^(−1) (x)+x))/x)dx sin(2tan^(−1) (x)+x)=sin(2tan^(−1) (x))cos(x)+cos(2tan^(−1) (x))sin(x) N.B:: (((sin(2tan^(−1) (x))= ((2x)/(1+x^2 )))),((cos(2tan^(−1) (x))= ((1−x^2 )/(1+x^2 )) )) ) Ω = ∫_0 ^( ∞) (((1/(1+x^2 ))(2xcos(x)+(1−x^2 )sin(x)))/x) Ω = ∫_0 ^( ∞) ((2cos(x))/(1+x^2 ))dx + ∫_0 ^( ∞) ((sin(x))/(x(1+x^2 )))dx − ∫_0 ^( ∞) ((xsin(x))/(1+x^2 ))dx Ω = 2∙(π/(2e)) + ∫_0 ^( ∞) ((sin(x))/(x(1+x^2 )))dx − ∫_0 ^( ∞) ((sin(x))/x)dx + ∫_0 ^( ∞) ((sin(x))/(x(1+x^2 )))dx Ω = (π/e) − (π/2) + 2Φ Φ = ∫_0 ^( ∞) ((sin(x))/(x(1+x^2 )))dx = (π/2)(1−e^(−1) ) Ω = (π/e) − (π/2) + π − (π/e) = (π/2) ★L𝛗rD ∅sE
$$ \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{sin}\left(\mathrm{2tan}^{−\mathrm{1}} \left(\mathrm{x}\right)+\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx} \\ $$$$\mathrm{sin}\left(\mathrm{2tan}^{−\mathrm{1}} \left(\mathrm{x}\right)+\mathrm{x}\right)=\mathrm{sin}\left(\mathrm{2tan}^{−\mathrm{1}} \left(\mathrm{x}\right)\right)\mathrm{cos}\left(\mathrm{x}\right)+\mathrm{cos}\left(\mathrm{2tan}^{−\mathrm{1}} \left(\mathrm{x}\right)\right)\mathrm{sin}\left(\mathrm{x}\right)\:\: \\ $$$$\mathrm{N}.\mathrm{B}::\:\begin{pmatrix}{\mathrm{sin}\left(\mathrm{2tan}^{−\mathrm{1}} \left(\mathrm{x}\right)\right)=\:\frac{\mathrm{2x}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\\{\mathrm{cos}\left(\mathrm{2tan}^{−\mathrm{1}} \left(\mathrm{x}\right)\right)=\:\frac{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:}\end{pmatrix} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\left(\mathrm{2xcos}\left(\mathrm{x}\right)+\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)\mathrm{sin}\left(\mathrm{x}\right)\right)}{\mathrm{x}} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{2cos}\left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:+\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx}\:−\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{xsin}\left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$$$\Omega\:=\:\mathrm{2}\centerdot\frac{\pi}{\mathrm{2e}}\:+\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx}\:−\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{x}}\mathrm{dx}\:+\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx} \\ $$$$\Omega\:=\:\frac{\pi}{\mathrm{e}}\:−\:\frac{\pi}{\mathrm{2}}\:+\:\mathrm{2}\Phi\: \\ $$$$\Phi\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{x}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx}\:=\:\frac{\pi}{\mathrm{2}}\left(\mathrm{1}−\mathrm{e}^{−\mathrm{1}} \right) \\ $$$$\Omega\:=\:\frac{\pi}{\mathrm{e}}\:−\:\frac{\pi}{\mathrm{2}}\:+\:\pi\:−\:\frac{\pi}{\mathrm{e}}\:=\:\frac{\pi}{\mathrm{2}} \\ $$$$\bigstar\boldsymbol{\mathrm{L}\phi\mathrm{rD}}\:\boldsymbol{\varnothing\mathrm{sE}} \\ $$

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