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 115366 by Bird last updated on 25/Sep/20

calculate ∫_(−1) ^2 (dx/(ch^2 x +sh^2 x))

$${calculate}\:\int_{−\mathrm{1}} ^{\mathrm{2}} \:\frac{{dx}}{{ch}^{\mathrm{2}} {x}\:+{sh}^{\mathrm{2}} {x}} \\ $$

Answered by MJS_new last updated on 25/Sep/20

∫(dx/(cosh^2 x +sinh^2 x))= [t=e^(2x) → dx=(dt/(2e^(2x) ))] =∫(dt/(t^2 +1))=arctan t = =arctan e^(2x) +C ∫_(−1) ^2 (dx/(cosh^2 x +sinh^2 x))=arctan e^4 −arctan e^(−2)

$$\int\frac{{dx}}{\mathrm{cosh}^{\mathrm{2}} \:{x}\:+\mathrm{sinh}^{\mathrm{2}} \:{x}}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{e}^{\mathrm{2}{x}} \:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{2e}^{\mathrm{2}{x}} }\right] \\ $$$$=\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}=\mathrm{arctan}\:{t}\:= \\ $$$$=\mathrm{arctan}\:\mathrm{e}^{\mathrm{2}{x}} \:+{C} \\ $$$$\underset{−\mathrm{1}} {\overset{\mathrm{2}} {\int}}\frac{{dx}}{\mathrm{cosh}^{\mathrm{2}} \:{x}\:+\mathrm{sinh}^{\mathrm{2}} \:{x}}=\mathrm{arctan}\:\mathrm{e}^{\mathrm{4}} \:−\mathrm{arctan}\:\mathrm{e}^{−\mathrm{2}} \\ $$

Commented by mathmax by abdo last updated on 25/Sep/20

thank you sir mjs

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{mjs} \\ $$

Answered by mathmax by abdo last updated on 25/Sep/20

let I =∫_(−1) ^2 (dx/(ch^2 x +sh^2 x)) ⇒I =∫_(−1) ^2 (dx/(((1+ch(2x))/2)+((ch(2x)−1)/2))) =∫_(−1) ^2 (dx/(ch(2x))) =2∫_(−1) ^2 (dx/(e^(2x) +e^(−2x) )) =_(e^(2x) =t) 2 ∫_e^(−2) ^e^(−4) (1/(t+t^(−1) ))×(dt/(2t)) =∫_e^(−2) ^e^(−4) (dt/(t^2 +1)) =arctan((1/e^4 ))−arctan((1/e^2 )) =(π/2) −arctan(e^4 )−((π/2)−arctan(e^2 )) =arctan(e^2 )−arctan(e^4 )

$$\mathrm{let}\:\mathrm{I}\:=\int_{−\mathrm{1}} ^{\mathrm{2}} \:\frac{\mathrm{dx}}{\mathrm{ch}^{\mathrm{2}} \mathrm{x}\:+\mathrm{sh}^{\mathrm{2}} \mathrm{x}}\:\Rightarrow\mathrm{I}\:=\int_{−\mathrm{1}} ^{\mathrm{2}} \:\frac{\mathrm{dx}}{\frac{\mathrm{1}+\mathrm{ch}\left(\mathrm{2x}\right)}{\mathrm{2}}+\frac{\mathrm{ch}\left(\mathrm{2x}\right)−\mathrm{1}}{\mathrm{2}}} \\ $$$$=\int_{−\mathrm{1}} ^{\mathrm{2}} \:\:\frac{\mathrm{dx}}{\mathrm{ch}\left(\mathrm{2x}\right)}\:=\mathrm{2}\int_{−\mathrm{1}} ^{\mathrm{2}} \:\frac{\mathrm{dx}}{\mathrm{e}^{\mathrm{2x}} +\mathrm{e}^{−\mathrm{2x}} }\:=_{\mathrm{e}^{\mathrm{2x}} \:=\mathrm{t}} \:\mathrm{2}\:\int_{\mathrm{e}^{−\mathrm{2}} } ^{\mathrm{e}^{−\mathrm{4}} } \:\:\:\frac{\mathrm{1}}{\mathrm{t}+\mathrm{t}^{−\mathrm{1}} }×\frac{\mathrm{dt}}{\mathrm{2t}} \\ $$$$=\int_{\mathrm{e}^{−\mathrm{2}} } ^{\mathrm{e}^{−\mathrm{4}} } \:\:\frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}}\:=\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{e}^{\mathrm{4}} }\right)−\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{e}^{\mathrm{2}} }\right) \\ $$$$=\frac{\pi}{\mathrm{2}}\:−\mathrm{arctan}\left(\mathrm{e}^{\mathrm{4}} \right)−\left(\frac{\pi}{\mathrm{2}}−\mathrm{arctan}\left(\mathrm{e}^{\mathrm{2}} \right)\right)\:=\mathrm{arctan}\left(\mathrm{e}^{\mathrm{2}} \right)−\mathrm{arctan}\left(\mathrm{e}^{\mathrm{4}} \right) \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com