Question Number 218779 by Spillover last updated on 15/Apr/25
Answered by breniam last updated on 15/Apr/25
![∫(1/(1+cosh x))dx=∫(1/(1+((e^x +e^(−x) )/2)))dx=∫(2/(e^x +2+e^(−x) ))dx= ∫(2/((e^(x/2) +e^(−(x/2)) )^2 ))dx={t=(x/2)}=4∫(dt/((e^t +e^(−t) )^2 ))={u=e^t }= =4∫(du/(u(u+(1/u))^2 ))=4∫((udu)/((u^2 +1)^2 ))={v=u^2 +1}= 2∫(dv/v^2 )=−(2/(e^x +1))+C lim_(x→∞) ((arcsinh(e^x ))/(e^x +1))=^H lim_(x→∞) ((e^x /( (√(e^(2x) +1))))/e^x )=0 ∫_0 ^∞ ((arcsinh(e^x ))/(1+cosh(x)))dx=−2∫_0 ^∞ ((1/(e^x +1)))′arcsinh(e^x )dx= =2∫_0 ^∞ ((e^x dx)/( (√(e^(2x) +1))(e^x +1)))={x^− =e^x }= 2∫_1 ^∞ (dx/( (√(x^2 +1))(x+1)))={x^− =arctan(x)}= 2∫_(π/4) ^(π/2) (dx/(cos(x)(tan(x)+1)))=2∫_(π/4) ^(π/2) (dx/(sin(x)+cos(x)))=2(√2)∫_(π/4) ^(π/2) (dx/(sin(x+(π/4))))=={x^− =x+(π/4)}= 2(√2)∫_(π/2) ^((3π)/4) (dx/(sin(x)))={x^− =cos(x)}= 2(√2)∫_(−((√2)/2)) ^0 (dx/(1−x^2 ))=−(√2)∫_(−((√2)/2)) ^0 [(1/(x−1))−(1/(x+1))]dx=−(√2)[ln∣((x−1)/(x+1))∣]_(−((√2)/2)) ^0 =(√2)ln(((1+((√2)/2))/(1−((√2)/2))))=2(√2)ln ((√2)(1+((√2)/2)))=2(√2)ln ((√2)+1)](https://www.tinkutara.com/question/Q218787.png)
$$ \\ $$$$\int\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cosh}\:{x}}\mathrm{d}{x}=\int\frac{\mathrm{1}}{\mathrm{1}+\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}}}\mathrm{d}{x}=\int\frac{\mathrm{2}}{{e}^{{x}} +\mathrm{2}+{e}^{−{x}} }\mathrm{d}{x}= \\ $$$$\int\frac{\mathrm{2}}{\left({e}^{\frac{{x}}{\mathrm{2}}} +{e}^{−\frac{{x}}{\mathrm{2}}} \right)^{\mathrm{2}} }\mathrm{d}{x}=\left\{{t}=\frac{{x}}{\mathrm{2}}\right\}=\mathrm{4}\int\frac{\mathrm{d}{t}}{\left({e}^{{t}} +{e}^{−{t}} \right)^{\mathrm{2}} }=\left\{{u}={e}^{{t}} \right\}= \\ $$$$=\mathrm{4}\int\frac{\mathrm{d}{u}}{{u}\left({u}+\frac{\mathrm{1}}{{u}}\right)^{\mathrm{2}} }=\mathrm{4}\int\frac{{u}\mathrm{d}{u}}{\left({u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }=\left\{{v}={u}^{\mathrm{2}} +\mathrm{1}\right\}= \\ $$$$\mathrm{2}\int\frac{\mathrm{d}{v}}{{v}^{\mathrm{2}} }=−\frac{\mathrm{2}}{{e}^{{x}} +\mathrm{1}}+{C} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{arcsinh}\left({e}^{{x}} \right)}{{e}^{{x}} +\mathrm{1}}\overset{{H}} {=}\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\frac{{e}^{{x}} }{\:\sqrt{{e}^{\mathrm{2}{x}} +\mathrm{1}}}}{{e}^{{x}} }=\mathrm{0} \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{arcsinh}\left({e}^{{x}} \right)}{\mathrm{1}+\mathrm{cosh}\left({x}\right)}\mathrm{d}{x}=−\mathrm{2}\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\frac{\mathrm{1}}{{e}^{{x}} +\mathrm{1}}\right)'\mathrm{arcsinh}\left({e}^{{x}} \right)\mathrm{d}{x}= \\ $$$$=\mathrm{2}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{e}^{{x}} \mathrm{d}{x}}{\:\sqrt{{e}^{\mathrm{2}{x}} +\mathrm{1}}\left({e}^{{x}} +\mathrm{1}\right)}=\left\{\overset{−} {{x}}={e}^{{x}} \right\}= \\ $$$$\mathrm{2}\underset{\mathrm{1}} {\overset{\infty} {\int}}\frac{\mathrm{d}{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\left({x}+\mathrm{1}\right)}=\left\{\overset{−} {{x}}=\mathrm{arctan}\left({x}\right)\right\}= \\ $$$$\mathrm{2}\underset{\frac{\pi}{\mathrm{4}}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{d}{x}}{\mathrm{cos}\left({x}\right)\left(\mathrm{tan}\left({x}\right)+\mathrm{1}\right)}=\mathrm{2}\underset{\frac{\pi}{\mathrm{4}}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{d}{x}}{\mathrm{sin}\left({x}\right)+\mathrm{cos}\left({x}\right)}=\mathrm{2}\sqrt{\mathrm{2}}\underset{\frac{\pi}{\mathrm{4}}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\frac{\mathrm{d}{x}}{\mathrm{sin}\left({x}+\frac{\pi}{\mathrm{4}}\right)}==\left\{\overset{−} {{x}}={x}+\frac{\pi}{\mathrm{4}}\right\}= \\ $$$$\mathrm{2}\sqrt{\mathrm{2}}\underset{\frac{\pi}{\mathrm{2}}} {\overset{\frac{\mathrm{3}\pi}{\mathrm{4}}} {\int}}\frac{\mathrm{d}{x}}{\mathrm{sin}\left({x}\right)}=\left\{\overset{−} {{x}}=\mathrm{cos}\left({x}\right)\right\}= \\ $$$$\mathrm{2}\sqrt{\mathrm{2}}\underset{−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} {\overset{\mathrm{0}} {\int}}\frac{\mathrm{d}{x}}{\mathrm{1}−{x}^{\mathrm{2}} }=−\sqrt{\mathrm{2}}\underset{−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} {\overset{\mathrm{0}} {\int}}\left[\frac{\mathrm{1}}{{x}−\mathrm{1}}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right]\mathrm{d}{x}=−\sqrt{\mathrm{2}}\left[\mathrm{ln}\mid\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\mid\right]_{−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} ^{\mathrm{0}} =\sqrt{\mathrm{2}}\mathrm{ln}\left(\frac{\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}{\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\right)=\mathrm{2}\sqrt{\mathrm{2}}\mathrm{ln}\:\left(\sqrt{\mathrm{2}}\left(\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\right)=\mathrm{2}\sqrt{\mathrm{2}}\mathrm{ln}\:\left(\sqrt{\mathrm{2}}+\mathrm{1}\right) \\ $$
Answered by Spillover last updated on 17/Apr/25
Answered by Spillover last updated on 17/Apr/25
