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Question Number 42487 by maxmathsup by imad last updated on 26/Aug/18
let f(x) = ∫_0 ^1 (dt/(x +ch(t))) 1) find a explicite form of f(x) 2) calculate ∫_0 ^1 (dt/((x+ch(t))^2 )) 3) find the value of ∫_0 ^1 (dt/(1+ch(t))) and ∫_0 ^1 (dt/(2+ch(t))) 4) find the value of ∫_0 ^1 (dt/((1+cht)^2 )) and ∫_0 ^1 (dt/((2+cht)^2 ))
$${let}\:\:{f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\:\frac{{dt}}{{x}\:+{ch}\left({t}\right)} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{explicite}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:\:{calculate}\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\left({x}+{ch}\left({t}\right)\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{dt}}{\mathrm{1}+{ch}\left({t}\right)}\:{and}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\mathrm{2}+{ch}\left({t}\right)} \\ $$$$\left.\mathrm{4}\right)\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{\left(\mathrm{1}+{cht}\right)^{\mathrm{2}} }\:\:{and}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{\left(\mathrm{2}+{cht}\right)^{\mathrm{2}} } \\ $$
Commented by maxmathsup by imad last updated on 28/Aug/18
1) we have f(x) = ∫_0 ^1 (dt/(x+((e^t +e^(−t) )/2))) =∫_0 ^1 ((2dt)/(2x +e^t +e^(−t) )) changement e^t =u give f(x) = ∫_1 ^e (2/(2x +u +(1/u))) (du/u) = ∫_1 ^e ((2du)/(2xu +u^2 +1)) = ∫_1 ^x ((2du)/(u^2 +2xu +1)) Δ^′ =x^2 −1 case 1 ∣x∣>1 ⇒u_1 =−x+(√(x^2 −1)) and u_2 =−x−(√(x^2 −1)) ⇒ F(u) = (2/(u^2 +2xu +1)) = (a/(u−u_1 )) +(b/(u−u_2 )) (=(2/((u−u_1 )(u−u_2 )))) a = (2/(u_1 −u_2 )) =(2/(2(√(x^2 −1)))) =(1/( (√(x^2 −1)))) b =(2/(u_2 −u_1 )) =(2/(−2(√(x^2 −1)))) =−(1/( (√(x^2 −1)))) ⇒ f(x) = ∫_1 ^e F(u)du =(1/( (√(x^2 −1)))) ∫_1 ^e ( (1/(u−u_1 )) −(1/(u−u_2 )))du =(1/( (√(x^2 −1))))[ln∣((u−u_1 )/(u−u_2 ))∣]_1 ^e =(1/( (√(x^2 −1)))){ln∣ ((e+x−(√(x^2 −1)))/(e+x+(√(x^2 −1))))∣ −ln∣((1+x−(√(x^2 −1)))/(1+x +(√(x^2 −1))))∣} case 2 ∣x∣<1 ⇒f(x)= ∫_1 ^e ((2du)/(u^2 +2xu +x^2 +1−x^2 )) =∫_1 ^e ((2du)/((u+x)^2 +1−x^2 )) =_(u+x =(√(1−x^2 ))α) ∫_((1+x)/( (√(1−x^2 )))) ^((e+x)/( (√(1−x^2 )))) ((2(√(1−x^2 ))dα)/((1−x^2 )(1+α^2 ))) =(2/( (√(1−x^2 )))) [ arctan(α)]_((1+x)/( (√(1−x^2 )))) ^((e+x)/( (√(1−x^2 )))) =(2/( (√(1−x^2 )))){ arctan(((e+x)/( (√(1−x^2 )))))−arctan(((1+x)/( (√(1−x^2 )))))} .
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{{x}+\frac{{e}^{{t}} \:+{e}^{−{t}} }{\mathrm{2}}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{2}{x}\:+{e}^{{t}} \:+{e}^{−{t}} }\:\:{changement}\:{e}^{{t}} \:={u}\:{give} \\ $$$${f}\left({x}\right)\:=\:\int_{\mathrm{1}} ^{{e}} \:\:\:\:\:\:\frac{\mathrm{2}}{\mathrm{2}{x}\:+{u}\:+\frac{\mathrm{1}}{{u}}}\:\frac{{du}}{{u}}\:=\:\int_{\mathrm{1}} ^{{e}} \:\:\:\:\:\frac{\mathrm{2}{du}}{\mathrm{2}{xu}\:+{u}^{\mathrm{2}} \:+\mathrm{1}}\:=\:\int_{\mathrm{1}} ^{{x}} \:\:\:\:\frac{\mathrm{2}{du}}{{u}^{\mathrm{2}} \:+\mathrm{2}{xu}\:+\mathrm{1}} \\ $$$$\Delta^{'} \:={x}^{\mathrm{2}} −\mathrm{1}\:\:\:\:{case}\:\mathrm{1}\:\:\mid{x}\mid>\mathrm{1}\:\Rightarrow{u}_{\mathrm{1}} =−{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\:\:{and}\:{u}_{\mathrm{2}} =−{x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\Rightarrow \\ $$$${F}\left({u}\right)\:=\:\frac{\mathrm{2}}{{u}^{\mathrm{2}} \:+\mathrm{2}{xu}\:+\mathrm{1}}\:=\:\frac{{a}}{{u}−{u}_{\mathrm{1}} }\:+\frac{{b}}{{u}−{u}_{\mathrm{2}} }\:\:\left(=\frac{\mathrm{2}}{\left({u}−{u}_{\mathrm{1}} \right)\left({u}−{u}_{\mathrm{2}} \right)}\right) \\ $$$${a}\:=\:\frac{\mathrm{2}}{{u}_{\mathrm{1}} −{u}_{\mathrm{2}} }\:=\frac{\mathrm{2}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:=\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}} \\ $$$${b}\:=\frac{\mathrm{2}}{{u}_{\mathrm{2}} −{u}_{\mathrm{1}} }\:=\frac{\mathrm{2}}{−\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:=−\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\:\int_{\mathrm{1}} ^{{e}} \:{F}\left({u}\right){du}\:=\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:\int_{\mathrm{1}} ^{{e}} \left(\:\:\frac{\mathrm{1}}{{u}−{u}_{\mathrm{1}} }\:−\frac{\mathrm{1}}{{u}−{u}_{\mathrm{2}} }\right){du} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\left[{ln}\mid\frac{{u}−{u}_{\mathrm{1}} }{{u}−{u}_{\mathrm{2}} }\mid\right]_{\mathrm{1}} ^{{e}} \:\:\:=\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\left\{{ln}\mid\:\:\:\frac{{e}+{x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{{e}+{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\mid\:−{ln}\mid\frac{\mathrm{1}+{x}−\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}{\mathrm{1}+{x}\:+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\mid\right\} \\ $$$${case}\:\mathrm{2}\:\:\mid{x}\mid<\mathrm{1}\:\Rightarrow{f}\left({x}\right)=\:\int_{\mathrm{1}} ^{{e}} \:\:\:\frac{\mathrm{2}{du}}{{u}^{\mathrm{2}} \:+\mathrm{2}{xu}\:+{x}^{\mathrm{2}} +\mathrm{1}−{x}^{\mathrm{2}} }\:=\int_{\mathrm{1}} ^{{e}} \:\:\:\frac{\mathrm{2}{du}}{\left({u}+{x}\right)^{\mathrm{2}} \:+\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$=_{{u}+{x}\:=\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\alpha} \:\:\:\:\:\:\:\int_{\frac{\mathrm{1}+{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}} ^{\frac{{e}+{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}} \:\:\:\:\:\:\:\frac{\mathrm{2}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{d}\alpha}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\left[\:{arctan}\left(\alpha\right)\right]_{\frac{\mathrm{1}+{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}} ^{\frac{{e}+{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}} \:\:\:\:=\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\left\{\:{arctan}\left(\frac{{e}+{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right)−{arctan}\left(\frac{\mathrm{1}+{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right)\right\}\:. \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 28/Aug/18
2) we have f^′ (x) = ∫_0 ^1 (∂/∂x)((1/(x+ch(t))))dt= −∫_0 ^1 (dt/((x+ch(t))^2 )) ⇒ ∫_0 ^1 (dt/((x +ch(t))^2 )) =−f^′ (x) and f is known..
$$\left.\mathrm{2}\right)\:{we}\:{have}\:\:{f}^{'} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\partial}{\partial{x}}\left(\frac{\mathrm{1}}{{x}+{ch}\left({t}\right)}\right){dt}=\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\left({x}+{ch}\left({t}\right)\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dt}}{\left({x}\:+{ch}\left({t}\right)\right)^{\mathrm{2}} }\:=−{f}^{'} \left({x}\right)\:\:{and}\:{f}\:{is}\:{known}.. \\ $$
Commented by maxmathsup by imad last updated on 28/Aug/18
3) ∫_0 ^1 (dt/(2+ch(t))) =f(2) = (1/( (√3))){ln∣((e+2−(√3))/(e+2+(√3)))∣ −ln∣ ((3−(√3))/(3+(√3)))∣ } .
$$\left.\mathrm{3}\right)\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{dt}}{\mathrm{2}+{ch}\left({t}\right)}\:={f}\left(\mathrm{2}\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left\{{ln}\mid\frac{{e}+\mathrm{2}−\sqrt{\mathrm{3}}}{{e}+\mathrm{2}+\sqrt{\mathrm{3}}}\mid\:−{ln}\mid\:\frac{\mathrm{3}−\sqrt{\mathrm{3}}}{\mathrm{3}+\sqrt{\mathrm{3}}}\mid\:\right\}\:. \\ $$

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