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Question-201291

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Question Number 201291 by sonukgindia last updated on 03/Dec/23
Answered by aleks041103 last updated on 03/Dec/23
I=∫_0 ^( 1) ((ln(1+x)dx)/(1+x^2 )) notice x=((1−t)/(1+t)), dx=−((2dt)/((1+t^2 ))) 1+x=((1−t)/(1+t))+1=(2/(1+t)) 1+x^2 =1+((1+t^2 −2t)/(1+t^2 +2t))=((2+2t^2 )/(1+t^2 +2t))=((2(1+t^2 ))/((1+t)^2 )) ⇒I=∫_1 ^( 0) ((ln(2)−ln(1+t))/((2(1+t^2 ))/((1+t)^2 )))(−((2dt)/((1+t)^2 )))= =∫_0 ^( 1) ((ln(2)−ln(1+t))/(1+t^2 ))dt=ln(2)∫_0 ^1 (dt/(1+t^2 ))−I ⇒2I=ln(2)∫_0 ^( 1) (dt/(1+t^2 ))=ln(2)arctan(1) ⇒I=((πln(2))/8)
$${I}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right){dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$${notice} \\ $$$${x}=\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}},\:{dx}=−\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)} \\ $$$$\mathrm{1}+{x}=\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}+\mathrm{1}=\frac{\mathrm{2}}{\mathrm{1}+{t}} \\ $$$$\mathrm{1}+{x}^{\mathrm{2}} =\mathrm{1}+\frac{\mathrm{1}+{t}^{\mathrm{2}} −\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} +\mathrm{2}{t}}=\frac{\mathrm{2}+\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} +\mathrm{2}{t}}=\frac{\mathrm{2}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{I}=\int_{\mathrm{1}} ^{\:\mathrm{0}} \frac{{ln}\left(\mathrm{2}\right)−{ln}\left(\mathrm{1}+{t}\right)}{\frac{\mathrm{2}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }}\left(−\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }\right)= \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{2}\right)−{ln}\left(\mathrm{1}+{t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}={ln}\left(\mathrm{2}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }−{I} \\ $$$$\Rightarrow\mathrm{2}{I}={ln}\left(\mathrm{2}\right)\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }={ln}\left(\mathrm{2}\right){arctan}\left(\mathrm{1}\right) \\ $$$$\Rightarrow{I}=\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{8}} \\ $$
Answered by Calculusboy last updated on 04/Dec/23
Solution: ∫_0 ^1 ((In(x+1))/(1+x^2 ))dx let x=tan𝛉 dx=sec^2 𝛉d𝛉 when x=1 𝛉=(𝛑/4) and when x=0 𝛉=0 I=∫_0 ^(𝛑/4) ((In(tan𝛉+1))/(1+tan^2 𝛉))sec^2 𝛉d𝛉 Nb: 1+tan^2 𝛉=sec^2 𝛉 I=∫_0 ^(𝛑/4) ((In(1+tan𝛉))/(sec^2 𝛉))sec^2 𝛉d𝛉 ⇔ I=∫_0 ^(𝛑/4) In(1+tan𝛉)d𝛉 let y=(𝛑/4)−𝛉 dy=−d𝛉 when 𝛉=(𝛑/4) y=0 and when 𝛉=0 y=(𝛑/4) I=∫_(𝛑/4) ^0 In [1+tan((𝛑/4)−y)](−dy) changing of variable I=∫_0 ^(𝛑/4) In[1+tan((𝛑/4)−𝛉)]d𝛉 ⇔ I=∫_0 ^(𝛑/4) In[1+((tan((𝛑/4))−tan𝛉)/(1+tan((𝛑/4))tan𝛉))]d𝛉 I=∫_0 ^(𝛑/4) In[1+((1−tan𝛉)/(1+tan𝛉))]d𝛉 ⇔ I=∫_0 ^(𝛑/4) In[((1+tan𝛉+1−tan𝛉)/(1+tan𝛉))]d𝛉 ⇔ I=∫_0 ^(𝛑/4) In[(2/(1+tan𝛉))]d𝛉 I=∫_0 ^(𝛑/4) [In2−In(1+tan𝛉)]d𝛉 ⇔ I=In(2)∫_0 ^(𝛑/4) d𝛉−∫_0 ^(𝛑/4) In(1+tan𝛉)d𝛉 I=In(2)[x]_0 ^(𝛑/4) −I 2I=In(2)[(𝛑/4)−0] 2I=(𝛑/4)In(2) I=(𝛑/8)In(2)
$$\boldsymbol{{Solution}}:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\boldsymbol{{In}}\left(\boldsymbol{{x}}+\mathrm{1}\right)}{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }\boldsymbol{{dx}}\:\:\:\:\:\:\boldsymbol{{let}}\:\boldsymbol{{x}}=\boldsymbol{{tan}\theta}\:\:\:\boldsymbol{{dx}}=\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{\theta{d}\theta} \\ $$$$\boldsymbol{{when}}\:\boldsymbol{{x}}=\mathrm{1}\:\:\boldsymbol{\theta}=\frac{\boldsymbol{\pi}}{\mathrm{4}}\:\:\boldsymbol{{and}}\:\boldsymbol{{when}}\:\boldsymbol{{x}}=\mathrm{0}\:\:\boldsymbol{\theta}=\mathrm{0} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \frac{\boldsymbol{{In}}\left(\boldsymbol{{tan}\theta}+\mathrm{1}\right)}{\mathrm{1}+\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{\theta}}\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{\theta{d}\theta}\:\:\:\boldsymbol{{Nb}}:\:\:\mathrm{1}+\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{\theta}=\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{\theta} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \:\frac{\boldsymbol{{In}}\left(\mathrm{1}+\boldsymbol{{tan}\theta}\right)}{\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{\theta}}\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{\theta{d}\theta}\:\:\:\Leftrightarrow\:\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \boldsymbol{{In}}\left(\mathrm{1}+\boldsymbol{{tan}\theta}\right)\boldsymbol{{d}\theta} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{y}}=\frac{\boldsymbol{\pi}}{\mathrm{4}}−\boldsymbol{\theta}\:\:\:\:\:\boldsymbol{{dy}}=−\boldsymbol{{d}\theta} \\ $$$$\boldsymbol{{when}}\:\boldsymbol{\theta}=\frac{\boldsymbol{\pi}}{\mathrm{4}}\:\:\boldsymbol{{y}}=\mathrm{0}\:\:\:\boldsymbol{{and}}\:\boldsymbol{{when}}\:\boldsymbol{\theta}=\mathrm{0}\:\:\:\boldsymbol{{y}}=\frac{\boldsymbol{\pi}}{\mathrm{4}} \\ $$$$\boldsymbol{{I}}=\int_{\frac{\boldsymbol{\pi}}{\mathrm{4}}} ^{\mathrm{0}} \boldsymbol{{In}}\:\left[\mathrm{1}+\boldsymbol{{tan}}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}−\boldsymbol{{y}}\right)\right]\left(−\boldsymbol{{dy}}\right) \\ $$$$\boldsymbol{{changing}}\:\boldsymbol{{of}}\:\boldsymbol{{variable}} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \:\boldsymbol{{In}}\left[\mathrm{1}+\boldsymbol{{tan}}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}−\boldsymbol{\theta}\right)\right]\boldsymbol{{d}\theta}\:\:\:\Leftrightarrow\:\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \:\boldsymbol{{In}}\left[\mathrm{1}+\frac{\boldsymbol{{tan}}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}\right)−\boldsymbol{{tan}\theta}}{\mathrm{1}+\boldsymbol{{tan}}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}\right)\boldsymbol{{tan}\theta}}\right]\boldsymbol{{d}\theta} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \boldsymbol{{In}}\left[\mathrm{1}+\frac{\mathrm{1}−\boldsymbol{{tan}\theta}}{\mathrm{1}+\boldsymbol{{tan}\theta}}\right]\boldsymbol{{d}\theta}\:\:\:\Leftrightarrow\:\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \boldsymbol{{In}}\left[\frac{\mathrm{1}+\boldsymbol{{tan}\theta}+\mathrm{1}−\boldsymbol{{tan}\theta}}{\mathrm{1}+\boldsymbol{{tan}\theta}}\right]\boldsymbol{{d}\theta}\:\:\:\:\:\Leftrightarrow\:\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \boldsymbol{{In}}\left[\frac{\mathrm{2}}{\mathrm{1}+\boldsymbol{{tan}\theta}}\right]\boldsymbol{{d}\theta} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \:\left[\boldsymbol{{In}}\mathrm{2}−\boldsymbol{{In}}\left(\mathrm{1}+\boldsymbol{{tan}\theta}\right)\right]\boldsymbol{{d}\theta}\:\:\:\Leftrightarrow\:\:\boldsymbol{{I}}=\boldsymbol{{In}}\left(\mathrm{2}\right)\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \boldsymbol{{d}\theta}−\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \boldsymbol{{In}}\left(\mathrm{1}+\boldsymbol{{tan}\theta}\right)\boldsymbol{{d}\theta} \\ $$$$\boldsymbol{{I}}=\boldsymbol{{In}}\left(\mathrm{2}\right)\left[\boldsymbol{{x}}\right]_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} −\boldsymbol{{I}} \\ $$$$\mathrm{2}\boldsymbol{{I}}=\boldsymbol{{In}}\left(\mathrm{2}\right)\left[\frac{\boldsymbol{\pi}}{\mathrm{4}}−\mathrm{0}\right] \\ $$$$\mathrm{2}\boldsymbol{{I}}=\frac{\boldsymbol{\pi}}{\mathrm{4}}\boldsymbol{{In}}\left(\mathrm{2}\right) \\ $$$$\boldsymbol{{I}}=\frac{\boldsymbol{\pi}}{\mathrm{8}}\boldsymbol{{In}}\left(\mathrm{2}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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