Question Number 109343 by 150505R last updated on 22/Aug/20
Commented by 150505R last updated on 22/Aug/20
$${please}\:{solve} \\ $$
Answered by mathmax by abdo last updated on 23/Aug/20
$$\mathrm{A}\:=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{2x}^{\mathrm{2}} −\mathrm{2x}\:+\mathrm{1}}\mathrm{dx}\:\Rightarrow\mathrm{A}\:=_{\mathrm{x}=\frac{\mathrm{t}}{\mathrm{2}}} \:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{t}}{\mathrm{2}}\right)}{\mathrm{2}\left(\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{4}}\right)−\mathrm{2}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)+\mathrm{1}}\frac{\mathrm{dt}}{\mathrm{2}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{2}−\mathrm{t}\right)}{\mathrm{2}\left\{\:\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{t}\:+\mathrm{1}\right\}}\:\mathrm{dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{2}−\mathrm{t}\right)}{\mathrm{t}^{\mathrm{2}} −\mathrm{2t}\:+\mathrm{2}}\:\mathrm{dt} \\ $$$$=_{\mathrm{t}=\mathrm{1}−\mathrm{u}} \:\:\:\:\:\:\int_{\mathrm{1}} ^{\mathrm{0}} \:\frac{\mathrm{ln}\left(\mathrm{2}−\mathrm{1}+\mathrm{u}\right)}{\left(\mathrm{1}−\mathrm{u}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}−\mathrm{u}\right)\:+\mathrm{2}}\left(−\mathrm{du}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{u}\right)}{\mathrm{u}^{\mathrm{2}} −\mathrm{2u}+\mathrm{1}−\mathrm{2}+\mathrm{2u}\:+\mathrm{2}}\:\mathrm{du}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{u}\right)}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }\:\mathrm{du} \\ $$$$=_{\mathrm{u}=\mathrm{tan}\theta} \:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{tan}\theta\right)}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta}\:\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta\right)\mathrm{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{sin}\theta}{\mathrm{cos}\theta}\right)\mathrm{d}\theta\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\mathrm{ln}\left(\mathrm{cos}\theta\:+\mathrm{sin}\theta\right)\mathrm{d}\theta−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\mathrm{ln}\left(\mathrm{cos}\theta\right)\mathrm{d}\theta \\ $$$$\mathrm{and}\:\mathrm{all}\:\mathrm{those}\:\mathrm{integrals}\:\mathrm{are}\:\mathrm{known}\:\mathrm{in}\:\mathrm{the}\:\mathrm{platform} \\ $$$$\left.\right)\left.\mathrm{i}\:\mathrm{dont}\:\mathrm{remember}\:\mathrm{the}\:\mathrm{values}\right)…. \\ $$
Answered by mathdave last updated on 22/Aug/20
![my solution here it goes I=∫_0 ^(1/2) ((ln(1−x))/((2x^2 −2x+1)))dx=(2/2)∫_0 ^(1/2) ((ln(1−x))/((2x^2 −2x+1)))dx=2∫_0 ^(1/2) ((ln(1−x))/((4x^2 −4x+2)))dx I=2∫_0 ^(1/2) ((ln(1−x))/((2x−1)^2 +1))dx (let 2x−1=y and dx=(dy/2)) I=(2/2)∫_(−1) ^0 ((ln(1−(((1+y)/2))))/(y^2 +1))dy=∫_(−1) ^0 ((ln(((1−y)/2)))/(1+y^2 ))dy let y=−t dy=−dt I=∫_1 ^0 ((ln(((1+t)/2)))/(1+t^2 ))×−dt=∫_0 ^1 ((ln(1+t))/(1+t^2 ))dt−∫_0 ^1 ((ln2)/(1+t^2 ))dt I=∫_0 ^1 ((ln(1+t))/(1+t^2 ))dt−ln2[tan^(−1) (t)]_0 ^1 =∫_0 ^1 ((ln(1+t))/(1+t^2 ))dt−(π/4)ln2 let tanx=t and dt=sec^2 xdx and 1+tan^2 x=sec^2 x I=∫_0 ^(π/4) ((ln(1+tanx))/(sec^2 x))×sec^2 xdx−(π/4)ln2=∫_0 ^(π/4) ln(1+((sinx)/(cosx)))dx−(π/4)ln2 I=∫_0 ^(π/4) ln(((sinx+cosx)/(cosx)))dx−(π/4)ln2=∫_0 ^(π/4) ln(cosx+sinx)dx−∫_0 ^(π/4) ln(cosx)dx−(π/4)ln2 since ∫_0 ^(π/2) ln(cosx+sinx)dx=G−(π/4)ln2 ∵∫_0 ^(π/4) ln(cosx+sinx)dx=(1/2)∫_0 ^(π/2) ln(cosx+sinx)dx=(1/2)(G−(π/4)ln2) and ∫_0 ^(π/4) ln(cosx)dx=(G/2)−(π/4)ln2 when uding this identity of ln(cosx)=−ln2+Σ_(n=1) ^∞ (((−1)^(n+1) cos(2nx))/n) from fourier series ∵I=(G/2)−(π/8)ln2−[(G/2)−(π/4)ln2]−(π/4)ln2 I=(G/2)−(G/2)−(π/8)ln2+(π/4)ln2−(π/4)ln2=−(π/8)ln2 ∵∫_0 ^(1/2) ((ln(1−x))/(2x^2 −2x+1))dx=−(π/8)ln2 by mathdave(22/08/2020)](https://www.tinkutara.com/question/Q109365.png)
$${my}\:{solution}\:{here}\:{it}\:{goes}\: \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{ln}\left(\mathrm{1}−{x}\right)}{\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\right)}{dx}=\frac{\mathrm{2}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{ln}\left(\mathrm{1}−{x}\right)}{\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\right)}{dx}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{ln}\left(\mathrm{1}−{x}\right)}{\left(\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{2}\right)}{dx} \\ $$$${I}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{ln}\left(\mathrm{1}−{x}\right)}{\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}{dx}\:\:\left({let}\:\:\mathrm{2}{x}−\mathrm{1}={y}\:\:\:{and}\:\:{dx}=\frac{{dy}}{\mathrm{2}}\right) \\ $$$${I}=\frac{\mathrm{2}}{\mathrm{2}}\int_{−\mathrm{1}} ^{\mathrm{0}} \frac{\mathrm{ln}\left(\mathrm{1}−\left(\frac{\mathrm{1}+{y}}{\mathrm{2}}\right)\right)}{{y}^{\mathrm{2}} +\mathrm{1}}{dy}=\int_{−\mathrm{1}} ^{\mathrm{0}} \frac{\mathrm{ln}\left(\frac{\mathrm{1}−{y}}{\mathrm{2}}\right)}{\mathrm{1}+{y}^{\mathrm{2}} }{dy} \\ $$$${let}\:\:{y}=−{t}\:\:\:{dy}=−{dt} \\ $$$${I}=\int_{\mathrm{1}} ^{\mathrm{0}} \frac{\mathrm{ln}\left(\frac{\mathrm{1}+{t}}{\mathrm{2}}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }×−{dt}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln2}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}−\mathrm{ln2}\left[\mathrm{tan}^{−\mathrm{1}} \left({t}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}−\frac{\pi}{\mathrm{4}}\mathrm{ln2} \\ $$$${let}\:\:\mathrm{tan}{x}={t}\:\:\:\:\:{and}\:\:\:{dt}=\mathrm{se}{c}^{\mathrm{2}} {xdx}\:\:\:{and}\: \\ $$$$\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {x}=\mathrm{sec}^{\mathrm{2}} {x} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{tan}{x}\right)}{\mathrm{sec}^{\mathrm{2}} {x}}×\mathrm{sec}^{\mathrm{2}} {xdx}−\frac{\pi}{\mathrm{4}}\mathrm{ln2}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{1}+\frac{\mathrm{sin}{x}}{\mathrm{cos}{x}}\right){dx}−\frac{\pi}{\mathrm{4}}\mathrm{ln2} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\frac{\mathrm{sin}{x}+\mathrm{cos}{x}}{\mathrm{cos}{x}}\right){dx}−\frac{\pi}{\mathrm{4}}\mathrm{ln2}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cos}{x}+\mathrm{sin}{x}\right){dx}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cos}{x}\right){dx}−\frac{\pi}{\mathrm{4}}\mathrm{ln2} \\ $$$${since}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{cos}{x}+\mathrm{sin}{x}\right){dx}={G}−\frac{\pi}{\mathrm{4}}\mathrm{ln2} \\ $$$$\because\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cos}{x}+\mathrm{sin}{x}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{cos}{x}+\mathrm{sin}{x}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}\left({G}−\frac{\pi}{\mathrm{4}}\mathrm{ln2}\right) \\ $$$${and}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cos}{x}\right){dx}=\frac{{G}}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\mathrm{ln2}\:\:\:\:\:\:{when}\:\:{uding}\:{this} \\ $$$${identity}\:{of}\: \\ $$$$\mathrm{ln}\left(\mathrm{cos}{x}\right)=−\mathrm{ln2}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \mathrm{cos}\left(\mathrm{2}{nx}\right)}{{n}}\:\:\:{from} \\ $$$${fourier}\:{series} \\ $$$$\because{I}=\frac{{G}}{\mathrm{2}}−\frac{\pi}{\mathrm{8}}\mathrm{ln2}−\left[\frac{{G}}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\mathrm{ln2}\right]−\frac{\pi}{\mathrm{4}}\mathrm{ln2} \\ $$$${I}=\frac{{G}}{\mathrm{2}}−\frac{{G}}{\mathrm{2}}−\frac{\pi}{\mathrm{8}}\mathrm{ln2}+\frac{\pi}{\mathrm{4}}\mathrm{ln2}−\frac{\pi}{\mathrm{4}}\mathrm{ln2}=−\frac{\pi}{\mathrm{8}}\mathrm{ln2} \\ $$$$\because\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{ln}\left(\mathrm{1}−{x}\right)}{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}{dx}=−\frac{\pi}{\mathrm{8}}\mathrm{ln2} \\ $$$${by}\:{mathdave}\left(\mathrm{22}/\mathrm{08}/\mathrm{2020}\right) \\ $$
Commented by mnjuly1970 last updated on 23/Aug/20
$${sir}.\:{tikalla}\:{mo}\:{barinam}\:{ti} \\ $$$${fargh}\:{sar}\:{man}\:{beshasham}… \\ $$$$ \\ $$