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answer-to-q25796-f-ind-the-value-off-x-0-pi-ln-1-xcos-d-with-0-lt-x-lt-1-f-x-0-pi-cos-1-xcos-1-d-pix-1-x-1-0-pi-1-xcos-1-d-and-by-the-changeent-tan-u-then-the

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Question Number 25821 by abdo imad last updated on 15/Dec/17
answer to q25796 f_ ind the value off(x)= ∫_0^ ^π_ ln(1+xcosθ)dθ with 0<x<1 ∂f/∂x= ∫_0 ^π cosθ(1+xcosθ)^(−1) dθ=πx^(−1) −x^(−1) ∫_0 ^π (1+xcosθ)^(−1) dθand by the changeent tanθ=u then the changement u=((1+x)(1+x)^(−1^ ) )^ .t.we find ∫_0 ^θ (1+xcosθ)^(−1) dθ =π(1−x^2_ )^(−1/2) so ∂f/∂x=πx^(−1) −πx^(−1) (1−x^(−1/2) ) so f(x)= π ln(x)−π∫^x t^(−1) (1−t^2 )^(−1/2) dt but ∫^x t^(−1) (1−t^2 )dt=−1. 2^(−1) ln((1+(1−x^2 )^(1/2) .(1−(1−x^2 )^(1/2) )^(−1) ) and f(x)=πln(x)+π.2^(−1) ln( (1+(1−x^2 )^(1/2) ((1−(1−x^2 )^(1/2) )^(−1) +β β=f(1)=∫_0 ^(π=) ln(1+cosθ)dθ=−πln(2) so ∫_0 ^π (1+xcosθ)dθ = πlnx +π2^(−1) ln((1+(1−x^2 )^(1/2^ ) )((1−(1−x^2 )^(1/2) )^(−1) −πln(2).
$${answer}\:{to}\:{q}\mathrm{25796}\:{f}_{} {ind}\:{the}\:{value}\:{off}\left({x}\right)=\:\int_{\mathrm{0}^{} } ^{\pi_{} } {ln}\left(\mathrm{1}+{xcos}\theta\right){d}\theta\:{with}\:\mathrm{0}<{x}<\mathrm{1}\:\:\:\partial{f}/\partial{x}=\:\int_{\mathrm{0}} ^{\pi} \:\:{cos}\theta\left(\mathrm{1}+{xcos}\theta\right)^{−\mathrm{1}} {d}\theta=\pi{x}^{−\mathrm{1}} −{x}^{−\mathrm{1}} \int_{\mathrm{0}} ^{\pi} \left(\mathrm{1}+{xcos}\theta\right)^{−\mathrm{1}} {d}\theta{and}\:{by}\:{the}\:{changeent}\:{tan}\theta={u}\:{then}\:{the}\:{changement}\:{u}=\left(\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+{x}\right)^{−\mathrm{1}^{} } \right)^{} \:.{t}.{we}\:{find}\:\:\int_{\mathrm{0}} ^{\theta} \left(\mathrm{1}+{xcos}\theta\right)^{−\mathrm{1}} {d}\theta\:\:=\pi\left(\mathrm{1}−{x}^{\mathrm{2}_{} } \right)^{−\mathrm{1}/\mathrm{2}} \\ $$$${so}\:\partial{f}/\partial{x}=\pi{x}^{−\mathrm{1}} −\pi{x}^{−\mathrm{1}} \left(\mathrm{1}−{x}^{−\mathrm{1}/\mathrm{2}} \right)\:{so}\:{f}\left({x}\right)=\:\pi\:{ln}\left({x}\right)−\pi\int^{{x}} \:{t}^{−\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{−\mathrm{1}/\mathrm{2}} {dt} \\ $$$${but}\:\int\:^{{x}} {t}^{−\mathrm{1}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right){dt}=−\mathrm{1}.\:\mathrm{2}^{−\mathrm{1}} \:{ln}\left(\left(\mathrm{1}+\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{2}} .\left(\mathrm{1}−\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{2}} \right)^{−\mathrm{1}} \right)\right. \\ $$$${and}\:{f}\left({x}\right)=\pi{ln}\left({x}\right)+\pi.\mathrm{2}^{−\mathrm{1}} {ln}\left(\:\left(\mathrm{1}+\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{2}} \left(\left(\mathrm{1}−\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{2}} \right)^{−\mathrm{1}} \:+\beta\right.\right.\right. \\ $$$$\beta={f}\left(\mathrm{1}\right)=\int_{\mathrm{0}} ^{\pi=} {ln}\left(\mathrm{1}+{cos}\theta\right){d}\theta=−\pi{ln}\left(\mathrm{2}\right)\:{so} \\ $$$$\int_{\mathrm{0}} ^{\pi} \left(\mathrm{1}+{xcos}\theta\right){d}\theta\:=\:\pi{lnx}\:+\pi\mathrm{2}^{−\mathrm{1}} {ln}\left(\left(\mathrm{1}+\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{2}^{} } \right)\left(\left(\mathrm{1}−\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{2}} \right)^{−\mathrm{1}} −\pi{ln}\left(\mathrm{2}\right).\right.\right. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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