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Question Number 32708 by abdo imad last updated on 31/Mar/18
let give f(x)=∫_0 ^(π/2) ((ln(1+xtant))/(tant))dt find a simple form of f(x) 2)calculate ∫_0 ^(π/2) ((ln(1+2tant))/(tant))dt .
$${let}\:{give}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{ln}\left(\mathrm{1}+{xtant}\right)}{{tant}}{dt} \\ $$$${find}\:{a}\:{simple}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{ln}\left(\mathrm{1}+\mathrm{2}{tant}\right)}{{tant}}{dt}\:. \\ $$
Commented by abdo imad last updated on 03/Apr/18
we have f^′ (x) = ∫_0 ^(π/2) (∂/∂x)( ((ln(1+xtant))/(tant)))dt = ∫_0 ^(π/2) ((tant)/((1+xtant)tant))dt = ∫_0 ^(π/2) (dt/(1+xtant)) = ∫_0 ^(π/2) (dt/(1+x((sint)/(cost)))) = ∫_0 ^(π/2) ((cost)/(cost +xsint)) dt .ch.tan((t/2))=u give f^, (x) = ∫_0 ^1 (((1−u^2 )/(1+u^2 ))/(((1−u^2 )/(1+u^2 )) +x (((2u))/(1+u^2 )))) ((2du)/(1+u^2 )) f^′ (x) =∫_0 ^1 ((2(1−u^2 ))/((1+u^2 )( 1−u^2 +2xu)))du = ∫_0 ^1 ((2(u^2 −1))/((1+u^2 )(u^2 −2xu −1)))du let decompose F(u) = ((2(u^2 −1))/((1+u^2 )( u^2 −2xu −1))) u^2 −2xu −1 ⇒Δ^′ =x^2 +1 ⇒ u_1 =x +(√(1+x^2 )) u_2 = x−(√(1+x^2 )) F(u) = (a/(u−u_1 )) +(b/(u−u_2 )) + ((cu +d)/(u^2 +1)) =((2(u^2 −1))/((u −u_1 )(u−u_2 )(u^2 +1))) a = ((2(u_1 ^2 −1))/((u_1 −u_2 )(u_1 ^2 +1))) = ((2( (x+(√(1+x^2 )))^2 −1))/(2(√(1+x^2 ( (x+(√(1+x^2 )))^2 +1))))) b = ((2( u_2 ^2 −1))/((u_2 −u_1 )(u_2 ^2 +1))) = ((2 ( (x−(√(1+x^2 )) )^2 −1))/(−2(√(2+x^2 ( ( x−(√(1+x^2 )^2 )) +1))))) be continued....
$${we}\:{have}\:{f}^{'} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{\partial}{\partial{x}}\left(\:\:\frac{{ln}\left(\mathrm{1}+{xtant}\right)}{{tant}}\right){dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{tant}}{\left(\mathrm{1}+{xtant}\right){tant}}{dt}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{dt}}{\mathrm{1}+{xtant}} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{dt}}{\mathrm{1}+{x}\frac{{sint}}{{cost}}}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{cost}}{{cost}\:+{xsint}}\:{dt}\:\:.{ch}.{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u} \\ $$$${give}\:{f}^{,} \left({x}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}{\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }\:+{x}\:\frac{\left(\mathrm{2}{u}\right)}{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$${f}^{'} \left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{2}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\:\mathrm{1}−{u}^{\mathrm{2}} \:+\mathrm{2}{xu}\right)}{du} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{2}\left({u}^{\mathrm{2}} −\mathrm{1}\right)}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left({u}^{\mathrm{2}} \:−\mathrm{2}{xu}\:−\mathrm{1}\right)}{du}\:{let}\:{decompose} \\ $$$${F}\left({u}\right)\:=\:\:\frac{\mathrm{2}\left({u}^{\mathrm{2}} −\mathrm{1}\right)}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\:{u}^{\mathrm{2}} \:−\mathrm{2}{xu}\:−\mathrm{1}\right)} \\ $$$${u}^{\mathrm{2}} \:−\mathrm{2}{xu}\:−\mathrm{1}\:\Rightarrow\Delta^{'} \:={x}^{\mathrm{2}} \:+\mathrm{1}\:\Rightarrow\:{u}_{\mathrm{1}} ={x}\:+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${u}_{\mathrm{2}} =\:{x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} \:}\:\: \\ $$$${F}\left({u}\right)\:=\:\:\frac{{a}}{{u}−{u}_{\mathrm{1}} }\:\:+\frac{{b}}{{u}−{u}_{\mathrm{2}} }\:\:+\:\frac{{cu}\:+{d}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:=\frac{\mathrm{2}\left({u}^{\mathrm{2}} −\mathrm{1}\right)}{\left({u}\:−{u}_{\mathrm{1}} \right)\left({u}−{u}_{\mathrm{2}} \right)\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$${a}\:=\:\frac{\mathrm{2}\left({u}_{\mathrm{1}} ^{\mathrm{2}} \:−\mathrm{1}\right)}{\left({u}_{\mathrm{1}} \:−{u}_{\mathrm{2}} \right)\left({u}_{\mathrm{1}} ^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\:\frac{\mathrm{2}\left(\:\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \:−\mathrm{1}\right)}{\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} \left(\:\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \:+\mathrm{1}\right)}} \\ $$$${b}\:=\:\frac{\mathrm{2}\left(\:{u}_{\mathrm{2}} ^{\mathrm{2}} \:−\mathrm{1}\right)}{\left({u}_{\mathrm{2}} \:−{u}_{\mathrm{1}} \right)\left({u}_{\mathrm{2}} ^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\:\:\frac{\mathrm{2}\:\left(\:\left({x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\right)^{\mathrm{2}} \:−\mathrm{1}\right)}{−\mathrm{2}\sqrt{\mathrm{2}+{x}^{\mathrm{2}} \left(\:\left(\:{x}−\sqrt{\left.\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} \:}\:+\mathrm{1}\right)\right.}} \\ $$$$\:{be}\:{continued}…. \\ $$

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