Question Number 59528 by Mr X pcx last updated on 11/May/19
$${let}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\mathrm{1}+{xch}\left({t}\right)}\:{with}\:{x}\:{real} \\ $$$$\left.\mathrm{1}\right)\:{determine}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right){find}\:{also}\:{g}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{\left(\mathrm{1}+{xch}\left({t}\right)\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\mathrm{1}+\mathrm{3}{ch}\left({t}\right)}\:{and}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{\left(\mathrm{1}+\mathrm{3}{ch}\left({t}\right)\right)^{\mathrm{2}} } \\ $$
Commented by MJS last updated on 11/May/19
$$\mathrm{same}\:\mathrm{path}\:\mathrm{as}\:\mathrm{before} \\ $$$$\int\frac{{dt}}{\mathrm{1}+{x}\mathrm{cosh}\:{t}}=\mathrm{2}\int\frac{\mathrm{e}^{{t}} }{{x}\mathrm{e}^{\mathrm{2}{t}} +\mathrm{2e}^{{t}} +{x}}{dt}=\frac{\mathrm{2}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\mathrm{arctan}\:\frac{{x}\mathrm{e}^{{t}} +\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:+{C} \\ $$$$\int\frac{{dt}}{\left(\mathrm{1}+{x}\mathrm{cosh}\:{t}\right)^{\mathrm{2}} }=\mathrm{4}\int\frac{\mathrm{e}^{\mathrm{2}{t}} }{\left({x}\mathrm{e}^{\mathrm{2}{t}} +\mathrm{2e}^{{t}} +{x}\right)^{\mathrm{2}} }{dt}=−\frac{\mathrm{2}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} }\mathrm{arctan}\:\frac{{x}\mathrm{e}^{{t}} +\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:−\frac{\mathrm{2}\left(\mathrm{e}^{{t}} +{x}\right)}{\left({x}\mathrm{e}^{\mathrm{2}{t}} +\mathrm{2e}^{{t}} +{x}\right)\left({x}^{\mathrm{2}} −\mathrm{1}\right)} \\ $$
Commented by maxmathsup by imad last updated on 11/May/19
$${thanks}\:{sir}\: \\ $$
Commented by maxmathsup by imad last updated on 12/May/19
![1) we have f(x) =∫_0 ^1 (dt/(1+x((e^t +e^(−t) )/2))) =∫_0 ^1 ((2dt)/(2+xe^t +x e^(−t) )) =_(e^t =u) ∫_1 ^e (2/(2 +xu +xu^(−1) )) (du/u) = ∫_1 ^e ((2du)/(2u+xu^2 +x)) =∫_1 ^e ((2du)/(xu^2 +2u +x)) poles of F(u)=(2/(xu^2 +2u +x)) xu^2 +2u +x =0 →Δ^′ =1−x^2 case 1 1−x^2 >0 ⇒∣x∣<1 ⇒u_1 =((−1+(√(1−x^2 )))/x) u_2 =((−1−(√(1−x^2 )))/x) (we suppose x≠0) ⇒F(u) =(2/(x(u−u_1 )(u−u_2 ))) =(2/(x(u_1 −u_2 )))( (1/(u−u_1 )) −(1/(u−u_2 ))) =(2/(x ((2(√(1−x^2 )))/x)))((1/(u−u_1 )) −(1/(u−u_2 )))=(1/( (√(1−x^2 ))))((1/(u−u_1 ))−(1/(u−u_2 ))) f(x) =∫_1 ^e F(u)du = (1/( (√(1−x^2 )))) [ln∣((u−u_1 )/(u−u_2 ))∣]_1 ^e =(1/( (√(1−x^2 )))){ln∣((e−u_1 )/(e−u_2 ))∣−ln∣((1−u_1 )/(1−u_2 ))∣} =(1/( (√(1−x^2 )))){ ln∣((e−((−1+(√(1−x^2 )))/x))/(e−((−1−(√(1−x^2 )))/x)))∣−ln∣((1−((−1+(√(1−x^2 )))/x))/(1−((−1−(√(1−x^2 )))/x)))∣} =(1/( (√(1−x^2 )))){ln∣((ex+1−(√(1−x^2 )))/(ex+1+(√(1−x^2 ))))∣−ln∣((x+1−(√(1−x^2 )))/(x+1+(√(1−x^2 ))))∣}](https://www.tinkutara.com/question/Q59598.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{\mathrm{1}+{x}\frac{{e}^{{t}} \:+{e}^{−{t}} }{\mathrm{2}}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{2}{dt}}{\mathrm{2}+{xe}^{{t}} \:+{x}\:{e}^{−{t}} } \\ $$$$=_{{e}^{{t}} \:={u}} \:\:\:\:\int_{\mathrm{1}} ^{{e}} \:\:\:\:\:\frac{\mathrm{2}}{\mathrm{2}\:+{xu}\:+{xu}^{−\mathrm{1}} }\:\frac{{du}}{{u}}\:=\:\int_{\mathrm{1}} ^{{e}} \:\:\:\:\frac{\mathrm{2}{du}}{\mathrm{2}{u}+{xu}^{\mathrm{2}} \:+{x}} \\ $$$$=\int_{\mathrm{1}} ^{{e}} \:\:\frac{\mathrm{2}{du}}{{xu}^{\mathrm{2}} \:+\mathrm{2}{u}\:+{x}}\:\:\:\:\:{poles}\:{of}\:{F}\left({u}\right)=\frac{\mathrm{2}}{{xu}^{\mathrm{2}} \:+\mathrm{2}{u}\:+{x}} \\ $$$${xu}^{\mathrm{2}} \:+\mathrm{2}{u}\:+{x}\:=\mathrm{0}\:\rightarrow\Delta^{'} =\mathrm{1}−{x}^{\mathrm{2}} \:\:\:{case}\:\mathrm{1}\:\:\:\:\mathrm{1}−{x}^{\mathrm{2}} >\mathrm{0}\:\Rightarrow\mid{x}\mid<\mathrm{1}\:\Rightarrow{u}_{\mathrm{1}} =\frac{−\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}} \\ $$$${u}_{\mathrm{2}} =\frac{−\mathrm{1}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}\:\:\:\left({we}\:{suppose}\:{x}\neq\mathrm{0}\right)\:\Rightarrow{F}\left({u}\right)\:=\frac{\mathrm{2}}{{x}\left({u}−{u}_{\mathrm{1}} \right)\left({u}−{u}_{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{2}}{{x}\left({u}_{\mathrm{1}} −{u}_{\mathrm{2}} \right)}\left(\:\:\frac{\mathrm{1}}{{u}−{u}_{\mathrm{1}} }\:−\frac{\mathrm{1}}{{u}−{u}_{\mathrm{2}} }\right)\:=\frac{\mathrm{2}}{{x}\:\frac{\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}}\left(\frac{\mathrm{1}}{{u}−{u}_{\mathrm{1}} }\:−\frac{\mathrm{1}}{{u}−{u}_{\mathrm{2}} }\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\left(\frac{\mathrm{1}}{{u}−{u}_{\mathrm{1}} }−\frac{\mathrm{1}}{{u}−{u}_{\mathrm{2}} }\right) \\ $$$${f}\left({x}\right)\:=\int_{\mathrm{1}} ^{{e}} \:{F}\left({u}\right){du}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\left[{ln}\mid\frac{{u}−{u}_{\mathrm{1}} }{{u}−{u}_{\mathrm{2}} }\mid\right]_{\mathrm{1}} ^{{e}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\left\{{ln}\mid\frac{{e}−{u}_{\mathrm{1}} }{{e}−{u}_{\mathrm{2}} }\mid−{ln}\mid\frac{\mathrm{1}−{u}_{\mathrm{1}} }{\mathrm{1}−{u}_{\mathrm{2}} }\mid\right\} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\left\{\:{ln}\mid\frac{{e}−\frac{−\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}}{{e}−\frac{−\mathrm{1}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}}\mid−{ln}\mid\frac{\mathrm{1}−\frac{−\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}}{\mathrm{1}−\frac{−\mathrm{1}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}}\mid\right\} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\left\{{ln}\mid\frac{{ex}+\mathrm{1}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{ex}+\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\mid−{ln}\mid\frac{{x}+\mathrm{1}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}+\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\mid\right\} \\ $$
Commented by maxmathsup by imad last updated on 12/May/19
![case 2 1−x^2 <0 ⇒x^2 −1>0⇒F(u) = (2/(xu^2 +2u +x)) =(2/(x(u^2 +2(u/x)+1))) =(2/(x{ u^2 +2(u/x) +(1/x^2 ) +1−(1/x^2 )))) =(2/(x{ (u+(1/x))^2 +((x^2 −1)/x^2 )})) =(2/(x{ (u+(1/x))^2 +(((√(x^2 −1))/(∣x∣)))^2 )) ⇒chang.u+(1/x) =((√(x^2 −1))/(∣x∣)) α give α =((xu +1)/x) ((∣x∣)/( (√(x^2 −1)))) f(x) =∫_1 ^e F(u)du = (2/x) ∫_(ξ(x)((x+1)/( (√(x^2 −1))))) ^(ξ(x)((ex+1)/( (√(x^2 −1))))) (1/(((x^2 −1)/(∣x∣))(1+α^2 ))) ((√(x^2 −1))/(∣x∣)) dα =(2/(x(√(x^2 −1)))) [arctan(α)]_(((x+1)/( (√(x^2 −1)))) ξ(x)) ^(((ex +1)/( (√(x^2 −1))))ξ(x)) =(2/(x(√(x^2 −1)))){ arctan(((ex+1)/( (√(x^2 −1)))) ξ(x)) −arctan(((x+1)/( (√(x^2 −1)))) ξ(x))} with ξ(x)=1 if x>0 and ξ(x) =−1 if x<0 .](https://www.tinkutara.com/question/Q59601.png)
$${case}\:\mathrm{2}\:\:\mathrm{1}−{x}^{\mathrm{2}} <\mathrm{0}\:\:\Rightarrow{x}^{\mathrm{2}} −\mathrm{1}>\mathrm{0}\Rightarrow{F}\left({u}\right)\:=\:\frac{\mathrm{2}}{{xu}^{\mathrm{2}} \:+\mathrm{2}{u}\:+{x}}\:=\frac{\mathrm{2}}{{x}\left({u}^{\mathrm{2}} \:+\mathrm{2}\frac{{u}}{{x}}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{2}}{{x}\left\{\:{u}^{\mathrm{2}} \:+\mathrm{2}\frac{{u}}{{x}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:+\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}\:=\frac{\mathrm{2}}{{x}\left\{\:\:\left({u}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} \:+\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} }\right\}} \\ $$$$=\frac{\mathrm{2}}{{x}\left\{\:\:\left({u}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} \:+\left(\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\mid{x}\mid}\right)^{\mathrm{2}} \right.}\:\Rightarrow{chang}.{u}+\frac{\mathrm{1}}{{x}}\:=\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\mid{x}\mid}\:\alpha\:\:{give}\:\alpha\:=\frac{{xu}\:+\mathrm{1}}{{x}}\:\frac{\mid{x}\mid}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}} \\ $$$${f}\left({x}\right)\:=\int_{\mathrm{1}} ^{{e}} \:{F}\left({u}\right){du}\:=\:\frac{\mathrm{2}}{{x}}\:\int_{\xi\left({x}\right)\frac{{x}+\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}} ^{\xi\left({x}\right)\frac{{ex}+\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}} \:\:\:\:\frac{\mathrm{1}}{\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\mid{x}\mid}\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)}\:\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\mid{x}\mid}\:{d}\alpha \\ $$$$=\frac{\mathrm{2}}{{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:\left[{arctan}\left(\alpha\right)\right]_{\frac{{x}+\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:\xi\left({x}\right)} ^{\frac{{ex}\:+\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\xi\left({x}\right)} \\ $$$$=\frac{\mathrm{2}}{{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\left\{\:{arctan}\left(\frac{{ex}+\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:\xi\left({x}\right)\right)\:−{arctan}\left(\frac{{x}+\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:\xi\left({x}\right)\right)\right\}\:{with} \\ $$$$\xi\left({x}\right)=\mathrm{1}\:{if}\:{x}>\mathrm{0}\:\:{and}\:\xi\left({x}\right)\:=−\mathrm{1}\:{if}\:{x}<\mathrm{0}\:. \\ $$
Commented by maxmathsup by imad last updated on 12/May/19
$$\left.\mathrm{3}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\mathrm{1}+\mathrm{3}\:{ch}\left({t}\right)}\:={f}\left(\mathrm{3}\right)\:\:\:\:\left({case}\:\mathrm{2}\right) \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{8}}}\left\{\:{arctan}\left(\frac{\mathrm{3}{e}+\mathrm{1}}{\:\sqrt{\mathrm{8}}}\right)−{arctan}\left(\frac{\mathrm{4}}{\:\sqrt{\mathrm{8}}}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}}\left\{\:{arctan}\left(\frac{\mathrm{3}{e}+\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\right)\:−{arctan}\left(\sqrt{\mathrm{2}}\right)\right\}\:. \\ $$