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