Integration Questions

Question Number 80921 by jagoll last updated on 08/Feb/20

$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{{xdx}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}\:=\:? \\$$

Commented by john santu last updated on 08/Feb/20

$${correction}\: \\$$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\frac{{xdx}}{\left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)^{{n}} }\:=\:\frac{\pi}{\mathrm{4}} \\$$

Commented by mathmax by abdo last updated on 10/Feb/20

$${A}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{xdx}}{{sinx}\:+{cosx}}\:\:{changement}\:{x}=\frac{\pi}{\mathrm{2}}−{t}\:{give} \\$$$${A}\:=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{\left(\frac{\pi}{\mathrm{2}}−{t}\right)\left(−{dt}\right)}{{cost}\:+{sint}}\:=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dt}}{{cost}\:+{sint}}\:−{A}\:\Rightarrow \\$$$$\mathrm{2}{A}=\frac{\pi}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{dt}}{{cost}\:+{sint}}\:\Rightarrow{A}=\frac{\pi}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{dt}}{{cost}\:+{sint}} \\$$$${changement}\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u}\:{give} \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{dt}}{{cost}\:+{sint}}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }+\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }}×\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{du}}{\mathrm{1}−{u}^{\mathrm{2}} \:+\mathrm{2}{u}} \\$$$$=−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{du}}{{u}^{\mathrm{2}} −\mathrm{2}{u}−\mathrm{1}}\:=−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{du}}{\left({u}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}}\:=−\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{du}}{\left({u}−\mathrm{1}−\sqrt{\mathrm{2}}\right)\left({u}−\mathrm{1}+\sqrt{\mathrm{2}}\right)} \\$$$$=−\frac{\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\frac{\mathrm{1}}{{u}−\mathrm{1}−\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{{u}−\mathrm{1}+\sqrt{\mathrm{2}}}\right\}{du} \\$$$$=−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\left[{ln}\mid\frac{{u}−\mathrm{1}−\sqrt{\mathrm{2}}}{{u}−\mathrm{1}+\sqrt{\mathrm{2}}}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} \:=−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\left(−{ln}\left(\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\sqrt{\mathrm{2}}−\mathrm{1}}\right)\:=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{ln}\left(\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} \right)\right. \\$$$$=\sqrt{\mathrm{2}}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:\Rightarrow{A}\:=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\$$