Question Number 31503 by abdo imad last updated on 09/Mar/18
$${find}\:\:\int_{\mathrm{2}} ^{\sqrt{\mathrm{5}}} \:\:\:\:\:\frac{{dt}}{{t}\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}}\:. \\ $$
Commented by abdo imad last updated on 16/Mar/18
![ch. t=ch(x) give I = ∫_(argch(2)) ^(argch((√5))) ((shxdx)/(ch(x)sh(x))) = ∫_(argch(2)) ^(argch((√5))) (dx/((e^x +e^(−x) )/2)) = 2 ∫_(argch2)) ^(argch((√5))) (dx/(e^x +e^(−x) )) let use the ch. e^x =t ⇒ I = 2 ∫_e^(argch(2)) ^e^(argch((√5))) (dt/(t(t +(1/t)))) = 2 ∫_e^(argch(2)) ^e^(argch((√5) )) (dt/(t^2 +1)) = 2 [ arctan(t)]_e^(argch(2)) ^e^(argch((√5))) but we know that argch(x)=ln(x +(√( x^2 −1_ )))⇒ argch(2)=ln(2+(√3)) and argch((√5))=ln((√5) +2) I= 2[arctan(t)]_(2+(√3)) ^(2+(√5)) =2( arctan(2+(√5))−arctan(2+(√3)))](https://www.tinkutara.com/question/Q31912.png)
$${ch}.\:{t}={ch}\left({x}\right)\:{give}\:\:{I}\:=\:\int_{{argch}\left(\mathrm{2}\right)} ^{{argch}\left(\sqrt{\mathrm{5}}\right)} \:\:\:\frac{{shxdx}}{{ch}\left({x}\right){sh}\left({x}\right)} \\ $$$$=\:\int_{{argch}\left(\mathrm{2}\right)} ^{{argch}\left(\sqrt{\mathrm{5}}\right)} \:\:\:\:\:\:\frac{{dx}}{\frac{{e}^{{x}} \:+{e}^{−{x}} }{\mathrm{2}}}\:=\:\mathrm{2}\:\int_{\left.{argch}\mathrm{2}\right)} ^{{argch}\left(\sqrt{\mathrm{5}}\right)} \:\:\frac{{dx}}{{e}^{{x}} \:+{e}^{−{x}} } \\ $$$${let}\:{use}\:{the}\:{ch}.\:{e}^{{x}} \:={t}\:\Rightarrow\:{I}\:=\:\mathrm{2}\:\int_{{e}^{{argch}\left(\mathrm{2}\right)} } ^{{e}^{{argch}\left(\sqrt{\mathrm{5}}\right)} } \:\:\:\:\:\:\frac{{dt}}{{t}\left({t}\:+\frac{\mathrm{1}}{{t}}\right)} \\ $$$$=\:\mathrm{2}\:\int_{{e}^{{argch}\left(\mathrm{2}\right)} } ^{{e}^{{argch}\left(\sqrt{\mathrm{5}}\:\right)} } \:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\:=\:\mathrm{2}\:\left[\:{arctan}\left({t}\right)\right]_{{e}^{{argch}\left(\mathrm{2}\right)} } ^{{e}^{{argch}\left(\sqrt{\mathrm{5}}\right)} } \:\:{but}\:{we}\:{know} \\ $$$${that}\:{argch}\left({x}\right)={ln}\left({x}\:+\sqrt{\:{x}^{\mathrm{2}} \:−\mathrm{1}_{\:} }\right)\Rightarrow \\ $$$${argch}\left(\mathrm{2}\right)={ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\:{and}\:{argch}\left(\sqrt{\mathrm{5}}\right)={ln}\left(\sqrt{\mathrm{5}}\:+\mathrm{2}\right) \\ $$$${I}=\:\mathrm{2}\left[{arctan}\left({t}\right)\right]_{\mathrm{2}+\sqrt{\mathrm{3}}} ^{\mathrm{2}+\sqrt{\mathrm{5}}} \:\:=\mathrm{2}\left(\:{arctan}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)−{arctan}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\right) \\ $$