Question Number 35060 by math khazana by abdo last updated on 14/May/18
$${calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{sinx}\:{ln}\left({cosx}\right){dx} \\ $$
Commented by math khazana by abdo last updated on 15/May/18
![let put I = ∫_0 ^(π/4) sinx ln(cosx)dx changement cosx =t give I = ∫_1 ^((√2)/2) (√(1−t^2 )) ln(t) −(dt/( (√(1−t^2 )))) = ∫_((√2)/2) ^1 ln(t)dt = [ t ln(t) −t_ ]_((√2)/2) ^1 = −1 −((√2)/2)ln(((√2)/2)) +((√2)/2) =((√2)/2) −1 −((√2)/2){ (1/2)ln(2) −ln(2)} =((√2)/2) −1 +(1/4)(√2) ln(2) .](https://www.tinkutara.com/question/Q35112.png)
$${let}\:{put}\:{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{sinx}\:{ln}\left({cosx}\right){dx}\:\:{changement} \\ $$$${cosx}\:={t}\:{give} \\ $$$${I}\:=\:\int_{\mathrm{1}} ^{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} \:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:\:{ln}\left({t}\right)\:−\frac{{dt}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }} \\ $$$$=\:\int_{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} ^{\mathrm{1}} \:\:{ln}\left({t}\right){dt}\:=\:\left[\:{t}\:{ln}\left({t}\right)\:−{t}_{} \right]_{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}} ^{\mathrm{1}} \: \\ $$$$=\:−\mathrm{1}\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}{ln}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:−\mathrm{1}\:\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left\{\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:−{ln}\left(\mathrm{2}\right)\right\} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:−\mathrm{1}\:\:+\frac{\mathrm{1}}{\mathrm{4}}\sqrt{\mathrm{2}}\:{ln}\left(\mathrm{2}\right)\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 15/May/18
![t=cosx dt=−sinxdx ∫_1 ^(1/( (√2))) −lnt dt I′=∫lnt dt =lnt×t−t so I=−∣tlnt−t∣_1 ^(1/( (√2))) =−[{(1/( (√2)))ln((1/( (√2))))−(1/( (√2)))}−{1.ln1−1}] =−[(1/( (√2)))(−(1/2)ln2)−(1/( (√2)))}−{−1}] =−[(−(1/(2(√2)))ln2−(1/( (√2)))}+1] =(1/(2(√2)))ln2+(1/( (√2)))−1 pls check](https://www.tinkutara.com/question/Q35077.png)
$${t}={cosx}\:\:{dt}=−{sinxdx} \\ $$$$\int_{\mathrm{1}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} −{lnt}\:{dt} \\ $$$${I}'=\int{lnt}\:{dt} \\ $$$$={lnt}×{t}−{t} \\ $$$${so}\:{I}=−\mid{tlnt}−{t}\mid_{\mathrm{1}} ^{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}} \\ $$$$=−\left[\left\{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right\}−\left\{\mathrm{1}.{ln}\mathrm{1}−\mathrm{1}\right\}\right] \\ $$$$\left.=−\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2}\right)−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right\}−\left\{−\mathrm{1}\right\}\right] \\ $$$$=−\left[\left(−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\mathrm{2}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right\}+\mathrm{1}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\mathrm{2}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\mathrm{1} \\ $$$${pls}\:{check} \\ $$
Commented by math khazana by abdo last updated on 15/May/18
$${your}\:{answer}\:{is}\:{correct}\:{sir}\:{Tanmay}…{thanks}… \\ $$