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 124594 by physicstutes last updated on 04/Dec/20

Show that      ∫_0 ^(ln 2) (1/(cosh(x + ln 4))) dx = 2 tan^(−1) ((4/(33)))

$$\mathrm{Show}\:\mathrm{that}\: \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\mathrm{ln}\:\mathrm{2}} \frac{\mathrm{1}}{\mathrm{cosh}\left({x}\:+\:\mathrm{ln}\:\mathrm{4}\right)}\:{dx}\:=\:\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{4}}{\mathrm{33}}\right) \\ $$

Commented by mohammad17 last updated on 04/Dec/20

let:y=x+ln4⇒dy=dx    x=0⇒y=ln4  ,x=ln2⇒y=ln2+ln4=ln8    ∫_(ln4) ^( ln8) (1/(cosh(y)))dy=∫_(ln4) ^( ln8) ((2/(e^y +e^(−y) )))dy=∫_(ln4) ^( ln8) (((2e^y )/(e^(2y) +1)))dy    =(2tan^(−1) (e^y ))_(ln4) ^(ln8)   =2tan^(−1) (8)−2tan^(−1) (4)    (mohammad Al−dolaimy)

$${let}:{y}={x}+{ln}\mathrm{4}\Rightarrow{dy}={dx} \\ $$$$ \\ $$$${x}=\mathrm{0}\Rightarrow{y}={ln}\mathrm{4}\:\:,{x}={ln}\mathrm{2}\Rightarrow{y}={ln}\mathrm{2}+{ln}\mathrm{4}={ln}\mathrm{8} \\ $$$$ \\ $$$$\int_{{ln}\mathrm{4}} ^{\:{ln}\mathrm{8}} \frac{\mathrm{1}}{{cosh}\left({y}\right)}{dy}=\int_{{ln}\mathrm{4}} ^{\:{ln}\mathrm{8}} \left(\frac{\mathrm{2}}{{e}^{{y}} +{e}^{−{y}} }\right){dy}=\int_{{ln}\mathrm{4}} ^{\:{ln}\mathrm{8}} \left(\frac{\mathrm{2}{e}^{{y}} }{{e}^{\mathrm{2}{y}} +\mathrm{1}}\right){dy} \\ $$$$ \\ $$$$=\left(\mathrm{2}{tan}^{−\mathrm{1}} \left({e}^{{y}} \right)\right)_{{ln}\mathrm{4}} ^{{ln}\mathrm{8}} \:\:=\mathrm{2}{tan}^{−\mathrm{1}} \left(\mathrm{8}\right)−\mathrm{2}{tan}^{−\mathrm{1}} \left(\mathrm{4}\right) \\ $$$$ \\ $$$$\left({mohammad}\:{Al}−{dolaimy}\right) \\ $$

Answered by Ar Brandon last updated on 04/Dec/20

I=∫_0 ^(ln2) (1/(cosh(x+ln4)))dx  x+ln4=u ⇒ dx=du  I=∫_(ln4) ^(ln8) (du/(cosh(u)))=∫_(ln4) ^(ln8) (du/((1/2)(e^u +e^(−u) )))     =2∫_(ln4) ^(ln8) ((e^u du)/(e^(2u) +1))=2∫_4 ^8 (dt/(t^2 +1))=2[tan^(−1) (t)]_4 ^8      =2[tan^(−1) (8)−tan^(−1) (4)]

$$\mathcal{I}=\int_{\mathrm{0}} ^{\mathrm{ln2}} \frac{\mathrm{1}}{\mathrm{cosh}\left(\mathrm{x}+\mathrm{ln4}\right)}\mathrm{dx} \\ $$$$\mathrm{x}+\mathrm{ln4}=\mathrm{u}\:\Rightarrow\:\mathrm{dx}=\mathrm{du} \\ $$$$\mathcal{I}=\int_{\mathrm{ln4}} ^{\mathrm{ln8}} \frac{\mathrm{du}}{\mathrm{cosh}\left(\mathrm{u}\right)}=\int_{\mathrm{ln4}} ^{\mathrm{ln8}} \frac{\mathrm{du}}{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{e}^{\mathrm{u}} +\mathrm{e}^{−\mathrm{u}} \right)} \\ $$$$\:\:\:=\mathrm{2}\int_{\mathrm{ln4}} ^{\mathrm{ln8}} \frac{\mathrm{e}^{\mathrm{u}} \mathrm{du}}{\mathrm{e}^{\mathrm{2u}} +\mathrm{1}}=\mathrm{2}\int_{\mathrm{4}} ^{\mathrm{8}} \frac{\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} +\mathrm{1}}=\mathrm{2}\left[\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{t}\right)\right]_{\mathrm{4}} ^{\mathrm{8}} \\ $$$$\:\:\:=\mathrm{2}\left[\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{8}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{4}\right)\right] \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com