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Question Number 124594 by physicstutes last updated on 04/Dec/20

$$\mathrm{Show}\mathrm{that}$$$${\int }_0^{\ln 2}\frac{1}{\cosh (x+\ln 4)}dx=2{\tan }^{-1}\left(\frac{4}{33}\right)$$

Commented by mohammad17 last updated on 04/Dec/20

$$let:y=x+ln4\Rightarrow dy=dx$$$$$$$$x=0\Rightarrow y=ln4,x=ln2\Rightarrow y=ln2+ln4=ln8$$$$$$$${\int }_{ln4}^{ln8}\frac{1}{cosh\left(y\right)}dy={\int }_{ln4}^{ln8}\left(\frac{2}{e^y+e^{-y}}\right)dy={\int }_{ln4}^{ln8}\left(\frac{2e^y}{e^{2y}+1}\right)dy$$$$$$$$={\left(2{tan}^{-1}\left(e^y\right)\right)}_{ln4}^{ln8}=2{tan}^{-1}\left(8\right)-2{tan}^{-1}\left(4\right)$$$$$$$$(mohammadAl-dolaimy)$$

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

$$\mathcal{I}={\int }_0^{\ln 2}\frac{1}{\cosh (\mathrm{x}+\ln 4)}\mathrm{dx}$$$$\mathrm{x}+\ln 4=\mathrm{u}\Rightarrow \mathrm{dx}=\mathrm{du}$$$$\mathcal{I}={\int }_{\ln 4}^{\ln 8}\frac{\mathrm{du}}{\cosh \left(\mathrm{u}\right)}={\int }_{\ln 4}^{\ln 8}\frac{\mathrm{du}}{\frac{1}{2}({\mathrm{e}}^{\mathrm{u}}+{\mathrm{e}}^{-\mathrm{u}})}$$$$=2{\int }_{\ln 4}^{\ln 8}\frac{{\mathrm{e}}^{\mathrm{u}}\mathrm{du}}{{\mathrm{e}}^{2\mathrm{u}}+1}=2{\int }_4^8\frac{\mathrm{dt}}{{\mathrm{t}}^2+1}=2{\left[{\tan }^{-1}\left(\mathrm{t}\right)\right]}_4^8$$$$=2[{\tan }^{-1}\left(8\right)-{\tan }^{-1}\left(4\right)]$$

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