Question Number 123154 by liberty last updated on 23/Nov/20 | ||
Answered by benjo_mathlover last updated on 23/Nov/20 | ||
$$\:{let}\:{t}=\mathrm{sin}\:\:{q}\:\Rightarrow{dt}\:=\:\mathrm{cos}\:{q}\:{dq} \\ $$$$\eta\:\left({x}\right)=\:\int\:\sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {q}}\:\left(\mathrm{cos}\:{q}\:{dq}\right) \\ $$$$\eta\:\left({x}\right)=\:\int\left(\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{2}{q}}{\mathrm{2}}\right)\:{dq}\: \\ $$$$\eta\:\left({x}\right)=\:\frac{{q}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{q}}{\mathrm{2}}\:+\:{c}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{sin}^{−\mathrm{1}} \left({t}\right)+{t}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{\mathrm{2}}\:+\:{c} \\ $$$${thus}\:\eta\:=\:\underset{\mathrm{1}} {\overset{\mathrm{cosh}\:\left({x}\right)} {\int}}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:{dt}\: \\ $$$$\:=\left[\:\frac{\mathrm{sin}^{−\mathrm{1}} \left({t}\right)+{t}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:}{\mathrm{2}}\right]_{\mathrm{1}} ^{\mathrm{cosh}\:\left({x}\right)} \\ $$$$\:=\:\left[\frac{\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{cosh}\:\left({x}\right)\right)+\mathrm{cosh}\:\left({x}\right)\mathrm{sinh}\:\left({x}\right)}{\mathrm{2}}\right]\:−\:\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$ | ||
Answered by Dwaipayan Shikari last updated on 23/Nov/20 | ||
$$\int{cos}\theta\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} \theta}\:{d}\theta \\ $$$$=\int{cos}^{\mathrm{2}} \theta{d}\theta=\int\frac{\mathrm{1}}{\mathrm{2}}+\frac{{cos}\mathrm{2}\theta}{\mathrm{2}}{d}\theta=\frac{\theta}{\mathrm{2}}+\frac{{sin}\mathrm{2}\theta}{\mathrm{4}}=\frac{{sin}^{−\mathrm{1}} {t}}{\mathrm{2}}+\frac{{t}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\int_{\mathrm{1}} ^{{coshx}} \sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:{dt}\:=\frac{{sin}^{−\mathrm{1}} \left(\frac{{e}^{{x}} −{e}^{−{x}} }{\mathrm{2}}\right)}{\mathrm{2}}+\frac{{coshx}\sqrt{\mathrm{1}−{cosh}^{\mathrm{2}} {x}}}{\mathrm{2}}−\frac{\pi}{\mathrm{4}} \\ $$ | ||
Answered by mathmax by abdo last updated on 23/Nov/20 | ||
$$\mathrm{f}\left(\mathrm{x}\right)=\int_{\mathrm{1}} ^{\frac{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} }{\mathrm{2}}} \sqrt{\mathrm{1}−\mathrm{t}^{\mathrm{2}} }\mathrm{dt}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)=_{\mathrm{t}=\mathrm{sin}\theta} \:\int_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{arcsin}\left(\frac{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} }{\mathrm{2}}\right)} \mathrm{cos}\theta\:\mathrm{cos}\theta\:\mathrm{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{arcsin}\left(\frac{\mathrm{e}^{\mathrm{x}} \:+\mathrm{e}^{−\mathrm{x}} }{\mathrm{2}}\right)} \left(\mathrm{1}+\mathrm{cos}\left(\mathrm{2}\theta\right)\right)\mathrm{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\theta\right]_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{arcsin}\left(\frac{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} }{\mathrm{2}}\right)} +\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{sin}\left(\mathrm{2}\theta\right)\right]_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{arcsin}\left(\frac{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} }{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{arcsin}\left(\frac{\mathrm{e}^{\mathrm{x}} \:+\mathrm{e}^{−\mathrm{x}} }{\mathrm{2}}\right)−\frac{\pi}{\mathrm{2}}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{sin}\theta\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta}\right]_{\frac{\pi}{\mathrm{2}}} ^{\mathrm{arcsin}\left(\frac{\mathrm{e}^{\mathrm{x}} \:+\mathrm{e}^{−\mathrm{x}} }{\mathrm{2}}\right)} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\:\mathrm{arcsin}\left(\frac{\mathrm{e}^{\mathrm{x}} \:+\mathrm{e}^{−\mathrm{x}} }{\mathrm{2}}\right)\right)−\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{e}^{\mathrm{x}} \:+\mathrm{e}^{−\mathrm{x}} }{\mathrm{2}}\sqrt{\mathrm{1}−\left(\frac{\mathrm{e}^{\mathrm{x}} +\mathrm{e}^{−\mathrm{x}} }{\mathrm{2}}\right)^{\mathrm{2}} }\right) \\ $$$$ \\ $$ | ||