UNKNOWN Questions

Question Number 53693 by gunawan last updated on 25/Jan/19

$$\:\underset{\:\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{{x}+\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{cos}\:{x}}\:{dx}\:= \\$$

Commented by maxmathsup by imad last updated on 25/Jan/19

$${let}\:{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{x}+{sinx}}{\mathrm{1}+{cosx}}\:\Rightarrow{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{x}}{\mathrm{1}+{cosx}}{dx}\:+\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{sinx}}{\mathrm{1}+{cosx}}{dx}\:{but} \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{sinx}}{\mathrm{1}+{cosx}}{dx}\:=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{\left({cosx}\right)^{,} }{\mathrm{1}+{cosx}}\:{dx}\:=−\left[{ln}\mid\mathrm{1}+{cosx}\mid\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =−\left(−{ln}\left(\mathrm{2}\right)\right)={ln}\left(\mathrm{2}\right) \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{x}}{\mathrm{1}+{cosx}}{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{x}}{{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right){dx}\:{but}\right. \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right){dx}\:=_{\frac{{x}}{\mathrm{2}}={t}} \:\:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\left(\mathrm{2}{t}\right)\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right)\left(\mathrm{2}\right){dt} \\$$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{t}\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){dt}\:\:\:{by}\:{parts}\:{u}={t}\:{and}\:{v}^{'} =\mathrm{1}+{tan}^{\mathrm{2}} {t}\:\Rightarrow \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{t}\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right){dt}\:=\left[{t}\:{tant}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{tant}\:{dt} \\$$$$=\frac{\pi}{\mathrm{4}}\:+\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{−{sint}}{{cost}}\:{dt}\:=\frac{\pi}{\mathrm{4}}\:+\left[{ln}\mid{cost}\mid\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:=\frac{\pi}{\mathrm{4}}\:+{ln}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\right)=\frac{\pi}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:\Rightarrow \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right){dx}\:=\pi−\mathrm{2}{ln}\left(\mathrm{2}\right)\:\Rightarrow \\$$$${I}\:={ln}\left(\mathrm{2}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\left(\pi−\mathrm{2}{ln}\left(\mathrm{2}\right)\right)\:\Rightarrow\:{I}\:=\frac{\pi}{\mathrm{2}}\:. \\$$$$\\$$

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Jan/19

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}}{\mathrm{1}+{cosx}}{dx}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\left(\mathrm{1}+{cosx}\right)}{\mathrm{1}+{cosx}} \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}×\frac{\mathrm{1}}{\mathrm{2}{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\left(\mathrm{1}+{cosx}\right)}{\mathrm{1}+{cosx}} \\$$$${now}... \\$$$$\int{xsec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}{dx} \\$$$${x}×\frac{{tan}\frac{{x}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{2}}}−\int\left[\frac{{dx}}{{dx}}\int{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}{dx}\right]{dx} \\$$$$=\mathrm{2}{xtan}\frac{{x}}{\mathrm{2}}−\int\frac{{tan}\frac{{x}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{2}}}{dx} \\$$$$=\mathrm{2}{xtan}\frac{{x}}{\mathrm{2}}−\mathrm{2}{lnsec}\frac{{x}}{\mathrm{2}} \\$$$${so} \\$$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {xsec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{d}\left(\mathrm{1}+{cosx}\right)}{\mathrm{1}+{cosx}} \\$$$$\frac{\mathrm{1}}{\mathrm{2}}\mid\mathrm{2}{xtan}\frac{{x}}{\mathrm{2}}−\mathrm{2}{lnsec}\frac{{x}}{\mathrm{2}}\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\mid{ln}\left(\mathrm{1}+{cosx}\right)\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\$$$$=\left\{\left(\frac{\pi}{\mathrm{2}}{tan}\frac{\pi}{\mathrm{4}}−{lnsec}\frac{\pi}{\mathrm{4}}\right)−\left(\mathrm{0}×{tan}\frac{\mathrm{0}}{\mathrm{2}}−{lnsec}\mathrm{0}\right)\right\}−\left\{{ln}\left(\mathrm{1}+\mathrm{0}\right)−{ln}\left(\mathrm{1}+\mathrm{1}\right)\right\} \\$$$$=\left\{\left(\frac{\pi}{\mathrm{2}}×\mathrm{1}−{ln}\sqrt{\mathrm{2}}\:\right)−\left(\mathrm{0}−\mathrm{0}\right)\right\}−\left\{\mathrm{0}−{ln}\mathrm{2}\right\} \\$$$$=\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2}+{ln}\mathrm{2} \\$$$$=\frac{\pi}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2} \\$$$${pls}\:{check}\:{steps}... \\$$$$\\$$