Question Number 61674 by behi83417@gmail.com last updated on 06/Jun/19
$$\mathrm{a}.\underset{\:\:\:\mathrm{0}} {\overset{\:\:\:\:\:\:\:\frac{\boldsymbol{\pi}}{\mathrm{4}}} {\int}}\sqrt{\mathrm{1}+\boldsymbol{\mathrm{tgx}}}\:\boldsymbol{\mathrm{dx}}=? \\ $$$$\mathrm{b}.\underset{\:\:\mathrm{0}} {\overset{\:\:\:\:\:\:\:\:\mathrm{1}} {\int}}\sqrt{\mathrm{1}+\boldsymbol{\mathrm{lnx}}}\:\boldsymbol{\mathrm{dx}}=? \\ $$
Commented by maxmathsup by imad last updated on 06/Jun/19
![b) we must have 1+ln(x)>0 βln(x)>β1 =ln((1/e)) βx>(1/e) so the fonction is not defined on]0,1] i think the Q is find β«_0 ^1 (β(1βln(x)))dx let use the chang (β(1βln(x)))=t β1βln(x) =t^2 βln(x)=1βt^2 β x =e^(1βt^2 ) β β«_0 ^1 (β(1βln(x)))dx = β«_(+β) ^1 t (β2t)e^(1βt^2 ) dt = 2β«_1 ^(+β) t^2 e^(1βt^2 ) dt =2e β«_1 ^(+β) t^2 e^(βt^2 ) dt by parts β«_1 ^(+β) t^2 e^(βt^2 ) dt =β(1/2)β«_1 ^(+β) t(β2t)e^(βt^2 ) dt =β(1/2){ [t e^(βt^2 ) ]_1 ^(+β) ββ«_1 ^(+β) e^(βt^2 ) dt} =β(1/2){ β(1/e) ββ«_1 ^(+β) e^(βt^2 ) dt} =(1/(2e)) +(1/2) β«_1 ^(+β) e^(βt^2 ) dt β«_0 ^β e^(βt^2 ) dt =β«_0 ^1 e^(βt^2 ) dt +β«_1 ^(+β) e^(βt^2 ) dt =((βΟ)/2) ββ«_1 ^(+β) e^(βt^2 ) dt=((βΟ)/2) ββ«_0 ^1 e^(βt^2 ) dt =((βΟ)/2) βΞ£_(n=0) ^β (((β1)^n )/(n!))β«_0 ^1 t^(2n) dt β β«_0 ^1 (β(1βln(x)))dx =(1/(2e)) +((βΟ)/4) β(1/2) Ξ£_(n=0) ^β (((β1)^n )/(n!(2n+1))) approximate value is β«_0 ^1 (β(1βln(x)))dx βΌ (1/(2e)) +((βΟ)/4) β(1/2){ 1 β(1/3) +(1/(10)) β(1/(21))} β β«_0 ^1 (β(1βln(x)))dx βΌ (1/(2e)) +((βΟ)/4) β(1/2) +(1/6) β(1/(20)) +(1/(42)) .](https://www.tinkutara.com/question/Q61698.png)
$$\left.{b}\right)\:{we}\:{must}\:{have}\:\mathrm{1}+{ln}\left({x}\right)>\mathrm{0}\:\Rightarrow{ln}\left({x}\right)>β\mathrm{1}\:={ln}\left(\frac{\mathrm{1}}{{e}}\right)\:\Rightarrow{x}>\frac{\mathrm{1}}{{e}}\:\:{so}\:{the}\:{fonction} \\ $$$$\left.{i}\left.{s}\:{not}\:{defined}\:{on}\right]\mathrm{0},\mathrm{1}\right]\:\:\:\:{i}\:\:{think}\:{the}\:{Q}\:{is}\:{find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}β{ln}\left({x}\right)}{dx} \\ $$$${let}\:{use}\:{the}\:{chang}\:\:\sqrt{\mathrm{1}β{ln}\left({x}\right)}={t}\:\Rightarrow\mathrm{1}β{ln}\left({x}\right)\:={t}^{\mathrm{2}} \:\Rightarrow{ln}\left({x}\right)=\mathrm{1}β{t}^{\mathrm{2}} \:\Rightarrow \\ $$$${x}\:={e}^{\mathrm{1}β{t}^{\mathrm{2}} } \:\:\:\Rightarrow\:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}β{ln}\left({x}\right)}{dx}\:=\:\int_{+\infty} ^{\mathrm{1}} {t}\:\:\left(β\mathrm{2}{t}\right){e}^{\mathrm{1}β{t}^{\mathrm{2}} } {dt} \\ $$$$=\:\mathrm{2}\int_{\mathrm{1}} ^{+\infty} \:{t}^{\mathrm{2}} \:{e}^{\mathrm{1}β{t}^{\mathrm{2}} } {dt}\:=\mathrm{2}{e}\:\int_{\mathrm{1}} ^{+\infty} \:\:{t}^{\mathrm{2}} \:{e}^{β{t}^{\mathrm{2}} } {dt}\:\:\:{by}\:{parts}\: \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\:{t}^{\mathrm{2}} \:{e}^{β{t}^{\mathrm{2}} } {dt}\:\:=β\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{+\infty} \:{t}\left(β\mathrm{2}{t}\right){e}^{β{t}^{\mathrm{2}} } {dt}\:=β\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\left[{t}\:{e}^{β{t}^{\mathrm{2}} } \right]_{\mathrm{1}} ^{+\infty} \:β\int_{\mathrm{1}} ^{+\infty} \:{e}^{β{t}^{\mathrm{2}} } {dt}\right\} \\ $$$$=β\frac{\mathrm{1}}{\mathrm{2}}\left\{\:β\frac{\mathrm{1}}{{e}}\:β\int_{\mathrm{1}} ^{+\infty} \:{e}^{β{t}^{\mathrm{2}} } {dt}\right\}\:=\frac{\mathrm{1}}{\mathrm{2}{e}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{+\infty} \:{e}^{β{t}^{\mathrm{2}} } {dt} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{β{t}^{\mathrm{2}} } {dt}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{β{t}^{\mathrm{2}} } {dt}\:+\int_{\mathrm{1}} ^{+\infty} \:{e}^{β{t}^{\mathrm{2}} } {dt}\:=\frac{\sqrt{\pi}}{\mathrm{2}}\:\Rightarrow\int_{\mathrm{1}} ^{+\infty} \:\:{e}^{β{t}^{\mathrm{2}} } {dt}=\frac{\sqrt{\pi}}{\mathrm{2}}\:β\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{β{t}^{\mathrm{2}} } {dt} \\ $$$$=\frac{\sqrt{\pi}}{\mathrm{2}}\:β\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(β\mathrm{1}\right)^{{n}} }{{n}!}\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{\mathrm{2}{n}} {dt}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}β{ln}\left({x}\right)}{dx}\:=\frac{\mathrm{1}}{\mathrm{2}{e}}\:+\frac{\sqrt{\pi}}{\mathrm{4}}\:β\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(β\mathrm{1}\right)^{{n}} }{{n}!\left(\mathrm{2}{n}+\mathrm{1}\right)} \\ $$$${approximate}\:{value}\:{is} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}β{ln}\left({x}\right)}{dx}\:\sim\:\:\frac{\mathrm{1}}{\mathrm{2}{e}}\:+\frac{\sqrt{\pi}}{\mathrm{4}}\:β\frac{\mathrm{1}}{\mathrm{2}}\left\{\:\:\mathrm{1}\:β\frac{\mathrm{1}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{10}}\:β\frac{\mathrm{1}}{\mathrm{21}}\right\}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}β{ln}\left({x}\right)}{dx}\:\sim\:\frac{\mathrm{1}}{\mathrm{2}{e}}\:+\frac{\sqrt{\pi}}{\mathrm{4}}\:β\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{6}}\:β\frac{\mathrm{1}}{\mathrm{20}}\:+\frac{\mathrm{1}}{\mathrm{42}}\:. \\ $$$$ \\ $$
Commented by behi83417@gmail.com last updated on 06/Jun/19
$$\mathrm{sir}\:\mathrm{proph}\:\mathrm{Abdo}!\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much} \\ $$$$\mathrm{for}\:\mathrm{so}\:\mathrm{hard}\:\mathrm{working}.\mathrm{god}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 06/Jun/19
$$\left.{a}\right){let}\:{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \sqrt{\mathrm{1}+{tanx}}{dx}\:{changeent}\:\sqrt{\mathrm{1}+{tanx}}={t}\:{give}\:\mathrm{1}+{tanx}\:={t}^{\mathrm{2}} \:\Rightarrow{tanx}={t}^{\mathrm{2}} β\mathrm{1}\:\Rightarrow \\ $$$${x}\:={arctan}\left({t}^{\mathrm{2}} β\mathrm{1}\right)\:\Rightarrow\:{I}\:=\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\:\frac{{t}}{\mathrm{1}+\left({t}^{\mathrm{2}} β\mathrm{1}\right)^{\mathrm{2}} }\left(\mathrm{2}{t}\right)\:{dt} \\ $$$$=\mathrm{2}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\:\:\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} β\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}}\:{dt}\:\:{let}\:{decompose}\:{F}\left({t}\right)\:=\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} β\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{2}} \\ $$$${t}^{\mathrm{4}} β\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{2}\:=\mathrm{0}\:\Rightarrow{u}^{\mathrm{2}} β\mathrm{2}{u}\:+\mathrm{2}\:=\mathrm{0}\:\:\:\left({u}={t}^{\mathrm{2}} \right) \\ $$$$\Delta^{'} \:=\mathrm{1}β\mathrm{2}=β\mathrm{1}<\mathrm{0}\:\:\Rightarrow\:{let}\:\lambda^{\mathrm{4}} \:=\mathrm{2}\:\Rightarrow{t}^{\mathrm{4}} β\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{2}\:={t}^{\mathrm{4}\:} +\lambda^{\mathrm{4}} β\mathrm{2}{t}^{\mathrm{2}} \\ $$$$=\left({t}^{\mathrm{2}} +\lambda^{\mathrm{2}} \right)^{\mathrm{2}} β\mathrm{2}\lambda^{\mathrm{2}} {t}^{\mathrm{2}} β\mathrm{2}{t}^{\mathrm{2}} \:=\left({t}^{\mathrm{2}} \:+\lambda^{\mathrm{2}} \right)^{\mathrm{2}} \:β\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right){t}^{\mathrm{2}} \\ $$$$\left.=\left({t}^{\mathrm{2}} \:+\lambda^{\mathrm{2}} β\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}{t}\right)\left({t}^{\mathrm{2}} \:+\lambda^{\mathrm{2}} +\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right.}\right){t}\right)\:\Rightarrow \\ $$$${F}\left({t}\right)\:=\frac{{at}\:+{b}}{{t}^{\mathrm{2}} β\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}{t}\:+\lambda^{\mathrm{2}} }\:+\frac{{ct}\:+{d}}{\left.{t}^{\mathrm{2}} +\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right.}\right){t}\:+\lambda^{\mathrm{2}} }\:\:\:\:\:\:\left(\lambda\:=\mathrm{2}^{\frac{\mathrm{1}}{\mathrm{4}}} \right) \\ $$$${F}\left(β{t}\right)={F}\left({t}\right)\:\Rightarrow{c}=β{a}\:{and}\:{d}={b}\:\Rightarrow \\ $$$${F}\left({t}\right)\:=\frac{{at}\:+{b}}{{t}^{\mathrm{2}} β\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}{t}\:+\lambda^{\mathrm{2}} }\:+\frac{β{at}\:+{b}}{\left.{t}^{\mathrm{2}} +\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right.}\right){t}\:+\lambda^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{0}\right)\:=\mathrm{0}\:=\frac{\mathrm{2}{b}}{\lambda^{\mathrm{2}} }\:\Rightarrow{b}=\mathrm{0} \\ $$$${F}\left(\mathrm{1}\right)\:=\:\mathrm{1}\:=\:\frac{{a}}{\lambda^{\mathrm{2}} \:+\mathrm{1}β\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right.}}\:β\frac{{a}}{\lambda^{\mathrm{2}} +\mathrm{1}+\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}}\:\Rightarrow \\ $$$$\mathrm{2}\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}{a}\:=\mathrm{1}\:\Rightarrow\:{a}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}}\:\Rightarrow \\ $$$$\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:{F}\left({t}\right){dt}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}}\left\{\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\:\:\frac{{tdt}}{{t}^{\mathrm{2}} β\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}{t}\:+\lambda^{\mathrm{2}} }\:β\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\:\:\frac{{tdt}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}{t}\:+\lambda^{\mathrm{2}} }\right\} \\ $$$$\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\:\frac{{tdt}}{{t}^{\mathrm{2}} β\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}{t}\:+\lambda^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\frac{\mathrm{2}{t}β\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}\:+\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}}{{t}^{\mathrm{2}} β\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}{t}\:+\lambda^{\mathrm{2}} }\:{dt} \\ $$$$\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\:\:\frac{{tdt}}{{t}^{\mathrm{2}} +\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}{t}\:+\lambda^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\frac{\mathrm{2}{t}\:+\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}\:β\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}{t}\:+\lambda^{\mathrm{2}} }\:{dt}\:\Rightarrow \\ $$$$\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:{F}\left({t}\right){dt}\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}}\left\{\:\:\left[{ln}\mid\frac{{t}^{\mathrm{2}} β\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}{t}\:+\mathrm{1}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}}\mid_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \right.\right. \\ $$$$+\frac{\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} β\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}{t}\:+\lambda^{\mathrm{2}} }\:+\frac{\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}{t}\:+\lambda^{\mathrm{2}} } \\ $$$$\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} β\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}{t}\:+\lambda^{\mathrm{2}} }\:=\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} β\mathrm{2}\:\frac{\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}}{\mathrm{2}}{t}\:\:\:+\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)+\lambda^{\mathrm{2}} β\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)} \\ $$$$=\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\:\:\:\:\:\:\frac{{dt}}{\left({t}β\frac{\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\lambda^{\mathrm{2}} β\mathrm{1}}{\mathrm{2}}}\:\:\:\:\:\left(\lambda^{\mathrm{2}} β\mathrm{1}=\sqrt{\mathrm{2}}β\mathrm{1}>\right) \\ $$$$=_{{t}β\frac{\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}}{\mathrm{2}}=\sqrt{\frac{\lambda^{\mathrm{2}} β\mathrm{1}}{\mathrm{2}}}{u}} \:\:\:\:\:\:\:\:\:\int_{\frac{\mathrm{2}β\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}}{\mathrm{2}\sqrt{\frac{\lambda^{\mathrm{2}} β\mathrm{1}}{\mathrm{2}}}}} ^{\frac{\mathrm{2}\sqrt{\mathrm{2}}β\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}}{\mathrm{2}\sqrt{\frac{\lambda^{\mathrm{2}} β\mathrm{1}}{\mathrm{2}}}}} \:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\frac{\lambda^{\mathrm{2}} β\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:\:\:\sqrt{\frac{\lambda^{\mathrm{2}} β\mathrm{1}}{\mathrm{2}}}{du} \\ $$$$=\sqrt{\frac{\mathrm{2}}{\lambda^{\mathrm{2}} β\mathrm{1}}}\:\left\{\:{arctan}\left(\frac{\mathrm{2}\sqrt{\mathrm{2}}β\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}}{\mathrm{2}\sqrt{\frac{\lambda^{\mathrm{2}} β\mathrm{1}}{\mathrm{2}}}}\right)β{arctan}\left(\frac{\mathrm{2}β\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}}{\mathrm{2}\sqrt{\frac{\lambda^{\mathrm{2}} β\mathrm{1}}{\mathrm{2}}}}\right)\right\} \\ $$$$=\sqrt{\frac{\mathrm{2}}{\:\sqrt{\mathrm{2}}β\mathrm{1}}}\left\{\:{arctan}\left(\frac{\mathrm{2}\sqrt{\mathrm{2}}β\sqrt{\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}}{\mathrm{2}\sqrt{\frac{\sqrt{\mathrm{2}}β\mathrm{1}}{\mathrm{2}}}}\right)β{arctan}\left(\frac{\mathrm{2}β\sqrt{\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)}}{\mathrm{2}\sqrt{\frac{\sqrt{\mathrm{2}}β\mathrm{1}}{\mathrm{2}}}}\right)\right\}{alsl} \\ $$$$\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} +\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}{t}\:+\lambda^{\mathrm{2}} }\:=_{{t}\:=β{x}} \:\:\:\:\:\int_{β\mathrm{1}} ^{β\sqrt{\mathrm{2}}} \:\:\:\:\:\frac{β{dx}}{{x}^{\mathrm{2}} β\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}{x}\:+\lambda^{\mathrm{2}} }\: \\ $$$$=β\int_{β\mathrm{1}} ^{β\sqrt{\mathrm{2}}} \:\:\:\:\:\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} β\sqrt{\mathrm{2}\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)}{x}+\lambda^{\mathrm{2}} }\:=….. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 06/Jun/19
$${you}\:{are}\:{welcome}\:{sir}. \\ $$
Answered by tanmay last updated on 06/Jun/19
![β«_0 ^(Ο/4) (β(1+tanx)) dx 1+tanx=a^2 sec^2 xΓ(dx/da)=2a dx=((2ada)/(1+(a^2 β1)^2 )) β«_1 ^(β2) ((2ada)/({1+(a^2 β1)^2 }))Γa β«_1 ^(β2) ((2a^2 )/(a^4 β2a^2 +2))da β«_1 ^(β2) (2/(a^2 β2+(2/a^2 )))da β«_1 ^(β2) ((1β((β2)/a^2 )+1+((β2)/a^2 ))/(a^2 +(2/a^2 )β2))da β«_1 ^(β2) ((d(a+((β2)/a)))/((a+((β2)/a))^2 β2(β2) β2))+β«_1 ^(β2) ((d(aβ((β2)/a)))/((aβ((β2)/a))^2 +2(β2) β2)) using formula β«(dx/(x^2 βa^2 )) andβ«(dx/(x^2 +a^2 )) =β«_1 ^(β2) ((d(a+((β2)/a)))/((a+((β2)/a))^2 β{(β((2(β2) +2))) }^2 )) +β«_1 ^(β2) ((d(aβ((β2)/a)))/((aβ((β2)/a))^2 +{(β(2(β2) β2)) }^2 )) =(1/(2{(β(2(β(2 )) +2)) }))Γβ£ln(((a+((β2)/a)β(β((2(β2) +2)) })/(a+((β2)/a)+(β(2(β2) +2)))))+(1/({(β(2(β2) β2)) }))tan^(β1) (((aβ((β2)/a))/( (β(2(β2) β2)))))β£_1 ^(β2) =[(1/(2{(β(2(β2) +2)) }))Γ{ln((((β2) +1β(β((2(β2) +2)))/( (β2) +1+(β((2(β2) +2))))}+(1/( (β(2(β2) β2))))tan^(β1) ((((β2) β1)/( (β(2(β2) β2)))))]β (1/(2{(β(2(β2) +2)) }))Γ{ln(((1+(β2) β(β(2(β2) +2)))/(1+(β2) +(β(2(β2) +2))))}β(1/( (β(2(β2) β2))))tan^(β1) (((1β(β2))/( (β(2(β2) β2))))) pls check](https://www.tinkutara.com/question/Q61705.png)
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \sqrt{\mathrm{1}+{tanx}}\:{dx} \\ $$$$\mathrm{1}+{tanx}={a}^{\mathrm{2}} \\ $$$${sec}^{\mathrm{2}} {x}Γ\frac{{dx}}{{da}}=\mathrm{2}{a} \\ $$$${dx}=\frac{\mathrm{2}{ada}}{\mathrm{1}+\left({a}^{\mathrm{2}} β\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\frac{\mathrm{2}{ada}}{\left\{\mathrm{1}+\left({a}^{\mathrm{2}} β\mathrm{1}\right)^{\mathrm{2}} \right\}}Γ{a} \\ $$$$\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \frac{\mathrm{2}{a}^{\mathrm{2}} }{{a}^{\mathrm{4}} β\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}}{da} \\ $$$$\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \frac{\mathrm{2}}{{a}^{\mathrm{2}} β\mathrm{2}+\frac{\mathrm{2}}{{a}^{\mathrm{2}} }}{da} \\ $$$$\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \frac{\mathrm{1}β\frac{\sqrt{\mathrm{2}}}{{a}^{\mathrm{2}} }+\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{{a}^{\mathrm{2}} }}{{a}^{\mathrm{2}} +\frac{\mathrm{2}}{{a}^{\mathrm{2}} }β\mathrm{2}}{da} \\ $$$$\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \frac{{d}\left({a}+\frac{\sqrt{\mathrm{2}}}{{a}}\right)}{\left({a}+\frac{\sqrt{\mathrm{2}}}{{a}}\right)^{\mathrm{2}} β\mathrm{2}\sqrt{\mathrm{2}}\:β\mathrm{2}}+\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \frac{{d}\left({a}β\frac{\sqrt{\mathrm{2}}}{{a}}\right)}{\left({a}β\frac{\sqrt{\mathrm{2}}}{{a}}\right)^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}\:β\mathrm{2}} \\ $$$${using}\:{formula}\:\int\frac{{dx}}{{x}^{\mathrm{2}} β{a}^{\mathrm{2}} }\:\:{and}\int\frac{{dx}}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \frac{{d}\left({a}+\frac{\sqrt{\mathrm{2}}}{{a}}\right)}{\left({a}+\frac{\sqrt{\mathrm{2}}}{{a}}\right)^{\mathrm{2}} β\left\{\sqrt{\left(\mathrm{2}\sqrt{\mathrm{2}}\:+\mathrm{2}\right)}\:\right\}^{\mathrm{2}} }\:+\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \frac{{d}\left({a}β\frac{\sqrt{\mathrm{2}}}{{a}}\right)}{\left({a}β\frac{\sqrt{\mathrm{2}}}{{a}}\right)^{\mathrm{2}} +\left\{\sqrt{\mathrm{2}\sqrt{\mathrm{2}}\:β\mathrm{2}}\:\right\}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left\{\sqrt{\mathrm{2}\sqrt{\mathrm{2}\:}\:+\mathrm{2}}\:\right\}}Γ\mid{ln}\left(\frac{\left.{a}+\frac{\sqrt{\mathrm{2}}}{{a}}β\sqrt{\left(\mathrm{2}\sqrt{\mathrm{2}}\:+\mathrm{2}\right.}\:\right\}}{{a}+\frac{\sqrt{\mathrm{2}}}{{a}}+\sqrt{\mathrm{2}\sqrt{\mathrm{2}}\:+\mathrm{2}}}\right)+\frac{\mathrm{1}}{\left\{\sqrt{\mathrm{2}\sqrt{\mathrm{2}}\:β\mathrm{2}}\:\right\}}{tan}^{β\mathrm{1}} \left(\frac{{a}β\frac{\sqrt{\mathrm{2}}}{{a}}}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{2}}\:β\mathrm{2}}}\right)\mid_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{2}\left\{\sqrt{\mathrm{2}\sqrt{\mathrm{2}}\:+\mathrm{2}}\:\right\}}Γ\left\{{ln}\left(\frac{\sqrt{\mathrm{2}}\:+\mathrm{1}β\sqrt{\left(\mathrm{2}\sqrt{\mathrm{2}}\:+\mathrm{2}\right.}}{\:\sqrt{\mathrm{2}}\:+\mathrm{1}+\sqrt{\left(\mathrm{2}\sqrt{\mathrm{2}}\:+\mathrm{2}\right.}}\right\}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{2}}\:β\mathrm{2}}}{tan}^{β\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}}\:β\mathrm{1}}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{2}}\:β\mathrm{2}}}\right)\right]β\right. \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\left\{\sqrt{\mathrm{2}\sqrt{\mathrm{2}}\:+\mathrm{2}}\:\right\}}Γ\left\{{ln}\left(\frac{\mathrm{1}+\sqrt{\mathrm{2}}\:β\sqrt{\mathrm{2}\sqrt{\mathrm{2}}\:+\mathrm{2}}}{\mathrm{1}+\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{2}\sqrt{\mathrm{2}}\:+\mathrm{2}}}\right\}β\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{2}}\:β\mathrm{2}}}{tan}^{β\mathrm{1}} \left(\frac{\mathrm{1}β\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}\sqrt{\mathrm{2}}\:β\mathrm{2}}}\right)\right. \\ $$$${pls}\:{check} \\ $$