Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 76143 by Master last updated on 24/Dec/19

Commented by mathmax by abdo last updated on 24/Dec/19

let I =∫_0 ^1 (√(4+x^2 ))dx  changement x=2sh(t) give  I =∫_0 ^(argsh((1/2)))  2ch(t)2ch(t)dt =4 ∫_0 ^(ln((1/2)+(√(1+(1/4))))) ch^2 (t)dt  =4 ∫_0 ^(ln((1/2)+((√5)/2))) ((ch(2t)−1)/2)dt =2 ∫_0 ^(ln(((1+(√5))/2))) (ch(2t)−1)dt  =[sh(2t)]_0 ^(ln(((1+(√5))/2))) −2ln(((1+(√5))/2))  =[((e^(2t) −e^(−2t) )/2)]_0 ^(ln(((1+(√5))/2))) −2ln(((1+(√5))/2))  =(1/2){  (((1+(√5))/2))^2 −(((1+(√5))/2))^(−2) }−2ln(((1+(√5))/2))

$${let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{4}+{x}^{\mathrm{2}} }{dx}\:\:{changement}\:{x}=\mathrm{2}{sh}\left({t}\right)\:{give} \\ $$$${I}\:=\int_{\mathrm{0}} ^{{argsh}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \:\mathrm{2}{ch}\left({t}\right)\mathrm{2}{ch}\left({t}\right){dt}\:=\mathrm{4}\:\int_{\mathrm{0}} ^{{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}}\right)} {ch}^{\mathrm{2}} \left({t}\right){dt} \\ $$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\right)} \frac{{ch}\left(\mathrm{2}{t}\right)−\mathrm{1}}{\mathrm{2}}{dt}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{{ln}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)} \left({ch}\left(\mathrm{2}{t}\right)−\mathrm{1}\right){dt} \\ $$$$=\left[{sh}\left(\mathrm{2}{t}\right)\right]_{\mathrm{0}} ^{{ln}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)} −\mathrm{2}{ln}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\ $$$$=\left[\frac{{e}^{\mathrm{2}{t}} −{e}^{−\mathrm{2}{t}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{{ln}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)} −\mathrm{2}{ln}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\:\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{−\mathrm{2}} \right\}−\mathrm{2}{ln}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right) \\ $$

Answered by john santuy last updated on 24/Dec/19

by letting x =2tan t →dx=2sec^2 t dt.   then ∫_0 ^(π/4) 2sec^2 t×2sect dt = ∫_0 ^(π/4) 4sec t d(tan t)  now can solve with integration by part

$${by}\:{letting}\:{x}\:=\mathrm{2}{tan}\:{t}\:\rightarrow{dx}=\mathrm{2}{sec}^{\mathrm{2}} {t}\:{dt}.\: \\ $$$${then}\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\mathrm{2}{sec}^{\mathrm{2}} {t}×\mathrm{2}{sect}\:{dt}\:=\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\mathrm{4}{sec}\:{t}\:{d}\left({tan}\:{t}\right) \\ $$$${now}\:{can}\:{solve}\:{with}\:{integration}\:{by}\:{part} \\ $$

Answered by benjo last updated on 24/Dec/19

Commented by Master last updated on 24/Dec/19

thanks

$$\mathrm{thanks} \\ $$

Commented by benjo last updated on 24/Dec/19

okay sir

$$\mathrm{okay}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com