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Question-201350

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Question Number 201350 by sonukgindia last updated on 05/Dec/23
Answered by Calculusboy last updated on 05/Dec/23
Solution: let y=(𝛑/4)βˆ’x dy=βˆ’dx when x=(𝛑/4) y=0 and when x=0 y=(𝛑/4) I=∫_(𝛑/4) ^0 In[1+tan((𝛑/4)βˆ’y)]βˆ’dy (change of variable) I=∫_0 ^(𝛑/4) In[1+tan((𝛑/4)βˆ’x)]dx ⇔ I=∫_0 ^(𝛑/4) In[1+((tan((𝛑/4))βˆ’tanx)/(1+tan((𝛑/4))tanx))]dx I=∫_0 ^(𝛑/4) In[1+((1βˆ’tanx)/(1+tanx))]dx ⇔ I=∫_0 ^(𝛑/4) In[((1+tanx+1βˆ’tanx)/(1+tanx))]dx I=∫_0 ^(𝛑/4) In[(2/(1+tanx))]dx ⇔ I=∫_0 ^(𝛑/4) [In(2)βˆ’In(1+tanx)]dx I=In(2)∫_0 ^(𝛑/4) dxβˆ’βˆ«_0 ^(𝛑/4) In(1+tanx)dx I=In(2)[x]_0 ^(𝛑/4) βˆ’I ⇔ 2I=In(2)[(𝛑/4)βˆ’0] I=(𝛑/4)Γ—(1/2)Γ—In(2) I=((𝛑In(2))/8)
$$\boldsymbol{{Solution}}:\:\boldsymbol{{let}}\:\boldsymbol{{y}}=\frac{\boldsymbol{\pi}}{\mathrm{4}}βˆ’\boldsymbol{{x}}\:\:\:\boldsymbol{{dy}}=βˆ’\boldsymbol{{dx}} \\ $$$$\boldsymbol{{when}}\:\boldsymbol{{x}}=\frac{\boldsymbol{\pi}}{\mathrm{4}}\:\:\boldsymbol{{y}}=\mathrm{0}\:\:\boldsymbol{{and}}\:\boldsymbol{{when}}\:\boldsymbol{{x}}=\mathrm{0}\:\boldsymbol{{y}}=\frac{\boldsymbol{\pi}}{\mathrm{4}} \\ $$$$\boldsymbol{{I}}=\int_{\frac{\boldsymbol{\pi}}{\mathrm{4}}} ^{\mathrm{0}} \boldsymbol{{In}}\left[\mathrm{1}+\boldsymbol{{tan}}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}βˆ’\boldsymbol{{y}}\right)\right]βˆ’\boldsymbol{{dy}}\:\:\:\:\left(\boldsymbol{{change}}\:\boldsymbol{{of}}\:\boldsymbol{{variable}}\right) \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \:\boldsymbol{{In}}\left[\mathrm{1}+\boldsymbol{{tan}}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}βˆ’\boldsymbol{{x}}\right)\right]\boldsymbol{{dx}}\:\:\Leftrightarrow\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \:\boldsymbol{{In}}\left[\mathrm{1}+\frac{\boldsymbol{{tan}}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}\right)βˆ’\boldsymbol{{tanx}}}{\mathrm{1}+\boldsymbol{{tan}}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}\right)\boldsymbol{{tanx}}}\right]\boldsymbol{{dx}} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \:\boldsymbol{{In}}\left[\mathrm{1}+\frac{\mathrm{1}βˆ’\boldsymbol{{tanx}}}{\mathrm{1}+\boldsymbol{{tanx}}}\right]\boldsymbol{{dx}}\:\:\Leftrightarrow\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \:\boldsymbol{{In}}\left[\frac{\mathrm{1}+\boldsymbol{{tanx}}+\mathrm{1}βˆ’\boldsymbol{{tanx}}}{\mathrm{1}+\boldsymbol{{tanx}}}\right]\boldsymbol{{dx}} \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \:\boldsymbol{{In}}\left[\frac{\mathrm{2}}{\mathrm{1}+\boldsymbol{{tanx}}}\right]\boldsymbol{{dx}}\:\Leftrightarrow\:\:\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \left[\boldsymbol{{In}}\left(\mathrm{2}\right)βˆ’\boldsymbol{{In}}\left(\mathrm{1}+\boldsymbol{{tanx}}\right)\right]\boldsymbol{{dx}} \\ $$$$\boldsymbol{{I}}=\boldsymbol{{In}}\left(\mathrm{2}\right)\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \boldsymbol{{dx}}βˆ’\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} \boldsymbol{{In}}\left(\mathrm{1}+\boldsymbol{{tanx}}\right)\boldsymbol{{dx}} \\ $$$$\boldsymbol{{I}}=\boldsymbol{{In}}\left(\mathrm{2}\right)\left[\boldsymbol{{x}}\right]_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{4}}} βˆ’\boldsymbol{{I}}\:\:\Leftrightarrow\:\:\mathrm{2}\boldsymbol{{I}}=\boldsymbol{{In}}\left(\mathrm{2}\right)\left[\frac{\boldsymbol{\pi}}{\mathrm{4}}βˆ’\mathrm{0}\right] \\ $$$$\boldsymbol{{I}}=\frac{\boldsymbol{\pi}}{\mathrm{4}}Γ—\frac{\mathrm{1}}{\mathrm{2}}Γ—\boldsymbol{{In}}\left(\mathrm{2}\right) \\ $$$$\boldsymbol{{I}}=\frac{\boldsymbol{\pi{In}}\left(\mathrm{2}\right)}{\mathrm{8}} \\ $$$$ \\ $$$$ \\ $$
Answered by Sutrisno last updated on 05/Dec/23
misal : I=∫_0 ^(Ο€/4) ln(1+tanx)dx (∫_0 ^a f(x)dx=∫_0 ^a f(aβˆ’x)dx) I=∫_0 ^(Ο€/4) ln(1+tan((Ο€/4)βˆ’x))dx I=∫_0 ^(Ο€/4) ln(1+((tan(Ο€/4)βˆ’tanx)/(1+tan(Ο€/4).tanx)))dx I=∫_0 ^(Ο€/4) ln(1+((1βˆ’tanx)/(1+tanx)))dx I=∫_0 ^(Ο€/4) ln((2/(1+tanx)))dx I=∫_0 ^(Ο€/4) ln(2)βˆ’βˆ«_0 ^(Ο€/4) ln(1+tanx)dx I=∫_0 ^(Ο€/4) ln(2)βˆ’I 2I=ln(2)x∣_0 ^(Ο€/4) I=((Ο€ln2)/8)
$${misal}\:: \\ $$$${I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}{ln}\left(\mathrm{1}+{tanx}\right){dx} \\ $$$$\:\left(\int_{\mathrm{0}} ^{{a}} {f}\left({x}\right){dx}=\int_{\mathrm{0}} ^{{a}} {f}\left({a}βˆ’{x}\right){dx}\right) \\ $$$${I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}{ln}\left(\mathrm{1}+{tan}\left(\frac{\pi}{\mathrm{4}}βˆ’{x}\right)\right){dx} \\ $$$${I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}{ln}\left(\mathrm{1}+\frac{{tan}\frac{\pi}{\mathrm{4}}βˆ’{tanx}}{\mathrm{1}+{tan}\frac{\pi}{\mathrm{4}}.{tanx}}\right){dx} \\ $$$${I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}{ln}\left(\mathrm{1}+\frac{\mathrm{1}βˆ’{tanx}}{\mathrm{1}+{tanx}}\right){dx} \\ $$$${I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}{ln}\left(\frac{\mathrm{2}}{\mathrm{1}+{tanx}}\right){dx} \\ $$$${I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}{ln}\left(\mathrm{2}\right)βˆ’\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}{ln}\left(\mathrm{1}+{tanx}\right){dx} \\ $$$${I}=\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}{ln}\left(\mathrm{2}\right)βˆ’{I} \\ $$$$\mathrm{2}{I}={ln}\left(\mathrm{2}\right){x}\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\mid}} \\ $$$${I}=\frac{\pi{ln}\mathrm{2}}{\mathrm{8}} \\ $$

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