Question Number 46667 by Necxx last updated on 29/Oct/18
$${integrate}\:\mathrm{ln}\:\left({cosx}+{sinx}\right){dx} \\ $$
Commented by maxmathsup by imad last updated on 29/Oct/18
![let f(x)=∫_0 ^x ln(cost +sint)dt we have f(x)= ∫_0 ^x ln((√2)cos(t−(π/4)))dt=(x/2)ln(2) +∫_0 ^x ln(cos(t−(π/4))dt but ∫_0 ^x ln(cos(t−(π/4)))dt =_(t−(π/4)=−u) ∫_(π/4) ^((π/4)−x) ln(cosu)du by parts = [u ln(cosu)]_(π/4) ^((π/4)−x) −∫_(π/4) ^((π/4)−x) u ((−sinu)/(cosu)) du = ∫_(π/4) ^((π/4)−x) u tan(u)( du changement tanu =t ) =∫_1 ^(tan((π/4)−x)) t arctant (dt/(1+t^2 )) =∫_1 ^(tan((π/4)−x)) ((t arctant)/(1+t^2 ))dt =[(1/2)ln(1+t^2 )arctant]_1 ^(tan((π/4)−x)) −∫_1 ^(tan((π/4)−x)) (1/2)ln(1+t^2 ) (dt/(1+t^2 )) =(1/2)((π/4)−x)ln(1+tan^2 ((π/4)−x))−(π/8)ln(2)−(1/2) ∫_1 ^(tan((π/4)−x)) ((ln(1+t^2 ))/(1+t^2 )) dt ...be continued...](Q46672.png)
$${let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} {ln}\left({cost}\:+{sint}\right){dt}\:\:{we}\:{have} \\ $$$${f}\left({x}\right)=\:\int_{\mathrm{0}} ^{{x}} {ln}\left(\sqrt{\mathrm{2}}{cos}\left({t}−\frac{\pi}{\mathrm{4}}\right)\right){dt}=\frac{{x}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:+\int_{\mathrm{0}} ^{{x}} {ln}\left({cos}\left({t}−\frac{\pi}{\mathrm{4}}\right){dt}\:{but}\right. \\ $$$$\int_{\mathrm{0}} ^{{x}} {ln}\left({cos}\left({t}−\frac{\pi}{\mathrm{4}}\right)\right){dt}\:=_{{t}−\frac{\pi}{\mathrm{4}}=−{u}} \:\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}−{x}} \:{ln}\left({cosu}\right){du}\:\:\:{by}\:{parts} \\ $$$$=\:\left[{u}\:{ln}\left({cosu}\right)\right]_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}−{x}} \:−\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}−{x}} \:{u}\:\:\frac{−{sinu}}{{cosu}}\:{du} \\ $$$$=\:\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}−{x}} \:\:{u}\:{tan}\left({u}\right)\left(\:{du}\:{changement}\:{tanu}\:={t}\:\right) \\ $$$$=\int_{\mathrm{1}} ^{{tan}\left(\frac{\pi}{\mathrm{4}}−{x}\right)} \:\:{t}\:{arctant}\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\int_{\mathrm{1}} ^{{tan}\left(\frac{\pi}{\mathrm{4}}−{x}\right)} \:\:\frac{{t}\:{arctant}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right){arctant}\right]_{\mathrm{1}} ^{{tan}\left(\frac{\pi}{\mathrm{4}}−{x}\right)} \:−\int_{\mathrm{1}} ^{{tan}\left(\frac{\pi}{\mathrm{4}}−{x}\right)} \:\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{4}}−{x}\right){ln}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}−{x}\right)\right)−\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right)−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{1}} ^{{tan}\left(\frac{\pi}{\mathrm{4}}−{x}\right)} \:\:\frac{{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt} \\ $$$$...{be}\:{continued}... \\ $$$$ \\ $$