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Question Number 42804 by maxmathsup by imad last updated on 02/Sep/18
calculate ∫_(1/2) ^1 (dx/( (√(4x^2 −1)) +(√(4x^2 +1))))
$${calculate}\:\:\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\:\:\:\:\frac{{dx}}{\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:−\mathrm{1}}\:+\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{1}}} \\ $$
Commented by maxmathsup by imad last updated on 03/Sep/18
let I = ∫_(1/2) ^1 (dx/( (√(4x^2 −1)) +(√(4x^2 +1)))) I =_(2x=u) ∫_1 ^2 (du/(2{ (√(u^2 −1)) +(√(u^2 +1})))) ⇒2I = ∫_1 ^2 (((√(u^2 +1))−(√(u^2 −1)))/2) du ⇒ 4I = ∫_1 ^2 (√(u^2 +1))du −∫_1 ^2 (√(u^2 −1))du but ∫_1 ^2 (√( 1+u^2 ))du =_(u=sh(x)) ∫_(ln(1+(√2))) ^(ln(2 +(√5))) ch(x)ch(x)dx =∫_(ln(1+(√2))) ^(ln(2+(√5))) ((1+ch(2x))/2)dx =(1/2) { ln(2+(√5))−ln(1+(√2))} +(1/4)[ sh(2x)]_(ln(1+(√2))) ^(ln−2+(√5))) =(1/2){ln(2+(√5))−ln(1+(√2))} +(1/4)[((e^(2x) −e^(−2x) )/2)]_(ln(1+(√2))) ^(ln(2+(√5))) =(1/2){ln(2+(√5))−ln(1+(√2))} +(1/8){(2+(√5))^2 −(2+(√5))^(−2) −(1+(√2))^2 +(1+(√2))^(−2) } also we have ∫_1 ^2 (√(u^2 −1))du =_(u=ch(x)) ∫_0 ^(ln(2+(√3))) sh(x)sh(x+dx =(1/2) ∫_0 ^(ln(2+(√3))) (ch(2x)−1)dx =−(1/2)ln(2+(√3)) +(1/4) [sh(2x)]_0 ^(ln(2+(√3))) =−(1/2)ln(2+(√3)) +(1/8)[ e^(2x) −e^(−2x) ]_0 ^(ln(2+(√3))) =−(1/2)ln(2+(√3)) +(1/8){ (2+(√3))^2 −(2+(√3))^(−2) } so the value of I is determined.
$${let}\:{I}\:=\:\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\:\:\frac{{dx}}{\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}}\:+\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$$${I}\:=_{\mathrm{2}{x}={u}} \:\:\:\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\frac{{du}}{\mathrm{2}\left\{\:\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}\:+\sqrt{\left.{u}^{\mathrm{2}} +\mathrm{1}\right\}}\right.}\:\Rightarrow\mathrm{2}{I}\:=\:\int_{\mathrm{1}} ^{\mathrm{2}} \:\:\frac{\sqrt{{u}^{\mathrm{2}} +\mathrm{1}}−\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{2}}\:\:{du}\:\Rightarrow \\ $$$$\mathrm{4}{I}\:=\:\int_{\mathrm{1}} ^{\mathrm{2}} \sqrt{{u}^{\mathrm{2}} +\mathrm{1}}{du}\:−\int_{\mathrm{1}} ^{\mathrm{2}} \sqrt{{u}^{\mathrm{2}} −\mathrm{1}}{du}\:\:\:{but} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{2}} \:\sqrt{\:\mathrm{1}+{u}^{\mathrm{2}} }{du}\:\:=_{{u}={sh}\left({x}\right)} \:\:\:\int_{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{{ln}\left(\mathrm{2}\:+\sqrt{\mathrm{5}}\right)} \:\:\:{ch}\left({x}\right){ch}\left({x}\right){dx}\:=\int_{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)} \:\frac{\mathrm{1}+{ch}\left(\mathrm{2}{x}\right)}{\mathrm{2}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\left\{\:{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\right\}\:+\frac{\mathrm{1}}{\mathrm{4}}\left[\:{sh}\left(\mathrm{2}{x}\right)\right]_{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{\left.{ln}−\mathrm{2}+\sqrt{\mathrm{5}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\right\}\:+\frac{\mathrm{1}}{\mathrm{4}}\left[\frac{{e}^{\mathrm{2}{x}} \:−{e}^{−\mathrm{2}{x}} }{\mathrm{2}}\right]_{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\right\}\:+\frac{\mathrm{1}}{\mathrm{8}}\left\{\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{\mathrm{2}} \:−\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)^{−\mathrm{2}} \:−\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \:+\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{−\mathrm{2}} \right\} \\ $$$${also}\:{we}\:{have}\:\int_{\mathrm{1}} ^{\mathrm{2}} \sqrt{{u}^{\mathrm{2}} −\mathrm{1}}{du}\:=_{{u}={ch}\left({x}\right)} \:\:\:\:\:\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)} {sh}\left({x}\right){sh}\left({x}+{dx}\right. \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)} \left({ch}\left(\mathrm{2}{x}\right)−\mathrm{1}\right){dx}\:=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:+\frac{\mathrm{1}}{\mathrm{4}}\:\left[{sh}\left(\mathrm{2}{x}\right)\right]_{\mathrm{0}} ^{{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:+\frac{\mathrm{1}}{\mathrm{8}}\left[\:{e}^{\mathrm{2}{x}} \:−{e}^{−\mathrm{2}{x}} \right]_{\mathrm{0}} ^{{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:+\frac{\mathrm{1}}{\mathrm{8}}\left\{\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:−\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{−\mathrm{2}} \right\}\:{so}\:{the}\:{value}\:{of}\:{I}\:{is}\:{determined}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Sep/18
=(1/2)∫_(1/2) ^1 (((4x^2 +1)−(4x^2 −1))/( (√(4x^2 +1)) +(√(4x^2 −1)) )) dx =(1/2)∫_(1/2) ^1 (√(4x^2 +1)) −(√(4x^2 −1)) dx =∫_(1/2) ^1 (√(x^2 +(1/2^2 ))) −(√(x^2 −(1/2^2 ))) dx now use formula ∫(√(x^2 +a^2 )) and∫(√(x^2 −a^2 )) dx ∣((x(√(x^2 +(1/4))))/2)+((1/2^2 )/2)ln(x+(√(x^2 +(1/4))) )−((x(√(x^2 −(1/4))))/2)+((1/2^2 )/2)ln(x+(√(x^2 −(1/4))) )∣_(1/2) ^1
$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\frac{\left(\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}\right)−\left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}\right)}{\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}\:+\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}}\:}\:{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{1}}\:\:−\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}}\:\:{dx} \\ $$$$=\int_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \:\sqrt{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }}\:\:−\sqrt{{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }}\:\:{dx} \\ $$$${now}\:{use}\:{formula}\:\:\int\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:\:{and}\int\sqrt{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }\:{dx} \\ $$$$\mid\frac{{x}\sqrt{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}}}{\mathrm{2}}+\frac{\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }}{\mathrm{2}}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}}\:\right)−\frac{{x}\sqrt{{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}}{\mathrm{2}}+\frac{\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }}{\mathrm{2}}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}\:\right)\mid_{\frac{\mathrm{1}}{\mathrm{2}}} ^{\mathrm{1}} \\ $$

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