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Question-144815




Question Number 144815 by mnjuly1970 last updated on 29/Jun/21
Answered by mindispower last updated on 30/Jun/21
=∫_0 ^∞ ((sin(x+(√(1+x^2 )))+sin((√(1+x^2 ))−x))/(2(√(1+x^2 )) ))dx  x=sh(t)  ∫_0 ^∞ ((sin(e^t )+sin(e^(−t) ))/(2ch(t))) ch(t)dt  =(1/2)∫_0 ^∞ sin(e^t )+sin(e^(−t) )dt,e^t =y firt,e^(−t) =x 2nd  =(1/2)∫_1 ^∞ ((sin(y))/y)dx+(1/2)∫_0 ^1 ((sin(x))/x)dx  =(1/2)∫_0 ^∞ ((sin(t))/t)dt=(π/4)
$$=\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)+{sin}\left(\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }−{x}\right)}{\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:}{dx} \\ $$$${x}={sh}\left({t}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({e}^{{t}} \right)+{sin}\left({e}^{−{t}} \right)}{\mathrm{2}{ch}\left({t}\right)}\:{ch}\left({t}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {sin}\left({e}^{{t}} \right)+{sin}\left({e}^{−{t}} \right){dt},{e}^{{t}} ={y}\:{firt},{e}^{−{t}} ={x}\:\mathrm{2}{nd} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\infty} \frac{{sin}\left({y}\right)}{{y}}{dx}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{sin}\left({x}\right)}{{x}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{sin}\left({t}\right)}{{t}}{dt}=\frac{\pi}{\mathrm{4}} \\ $$

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