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Question Number 84242 by Rio Michael last updated on 10/Mar/20

$$\underset{\mathrm{0}} {\overset{\mathrm{ln2}} {\int}}\frac{\mathrm{1}}{\mathrm{cosh}\left({x}\:+\:\mathrm{ln4}\right)}{dx} \\$$

Commented by mathmax by abdo last updated on 10/Mar/20

$${I}\:=\int_{\mathrm{0}} ^{{ln}\left(\mathrm{2}\right)} \:\frac{\mathrm{1}}{{ch}\left({x}+{ln}\mathrm{4}\right)}{dx}\:\Rightarrow{I}\:=_{{x}+{ln}\mathrm{4}={t}} \:\:\int_{\mathrm{2}{ln}\left(\mathrm{2}\right)} ^{\mathrm{3}{ln}\left(\mathrm{2}\right)} \:\:\frac{{dt}}{{ch}\left({t}\right)} \\$$$$=\mathrm{2}\:\int_{\mathrm{2}{ln}\left(\mathrm{2}\right)} ^{\mathrm{3}{ln}\left(\mathrm{2}\right)} \:\frac{{dt}}{{e}^{{t}} \:+{e}^{−{t}} }\:=_{{e}^{{t}} ={u}} \:\:\mathrm{2}\:\int_{\mathrm{4}} ^{\mathrm{8}} \:\frac{{du}}{{u}\left({u}+{u}^{−\mathrm{1}} \right)}\:=\mathrm{2}\int_{\mathrm{4}} ^{\mathrm{8}} \:\frac{{du}}{{u}^{\mathrm{2}} \:+\mathrm{1}} \\$$$$=\mathrm{2}\left[{arctanu}\right]_{\mathrm{4}} ^{\mathrm{8}} \:=\mathrm{2}\left({arctan}\left(\mathrm{8}\right)−{arctan}\left(\mathrm{4}\right)\right) \\$$

Answered by TANMAY PANACEA last updated on 10/Mar/20

$${cosh}\left({x}+{ln}\mathrm{4}\right)=\frac{{e}^{{x}+{ln}\mathrm{4}} +{e}^{−\left({x}+{ln}\mathrm{4}\right)} }{\mathrm{2}} \\$$$$=\frac{{e}^{{x}} .{e}^{{ln}\mathrm{4}} +{e}^{−{x}} .{e}^{−{ln}\mathrm{4}} }{\mathrm{2}} \\$$$${t}={e}^{{ln}\mathrm{4}} \rightarrow{lnt}={ln}\mathrm{4}\rightarrow\rightarrow{t}=\mathrm{4} \\$$$$=\frac{{e}^{{x}} ×\mathrm{4}+{e}^{−{x}} ×\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{2}}=\frac{\mathrm{16}{e}^{\mathrm{2}{x}} +\mathrm{1}}{\mathrm{8}{e}^{{x}} } \\$$$$\int_{\mathrm{0}} ^{{ln}\mathrm{2}} \frac{\mathrm{8}{e}^{{x}} {dx}}{\mathrm{16}{e}^{\mathrm{2}{x}} +\mathrm{1}} \\$$$$\frac{\mathrm{8}}{\mathrm{16}}\int_{\mathrm{0}} ^{{ln}\mathrm{2}} \frac{{e}^{{x}} {dx}}{\left({e}^{{x}} \right)+\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} }\rightarrow\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{{ln}\mathrm{2}} \frac{{d}\left({e}^{{x}} \right)}{\left({e}^{{x}} \right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} } \\$$$$\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{4}}}\mid{tan}^{−\mathrm{1}} \left(\frac{{e}^{{x}} }{\frac{\mathrm{1}}{\mathrm{4}}}\right)\mid_{\mathrm{0}} ^{{ln}\mathrm{2}} \\$$$$\mathrm{2}\left({tan}^{−\mathrm{1}} \mathrm{8}−{tan}^{−\mathrm{1}} \mathrm{4}\right) \\$$$$\\$$$$\\$$

Commented by Rio Michael last updated on 10/Mar/20

$$\mathrm{thanks} \\$$