Integration Questions

Question Number 83805 by M±th+et£s last updated on 06/Mar/20

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}^{\mathrm{2}} \left({x}\right)}{{sin}\left({x}\right)+{cos}\left({x}\right)}\:{dx} \\$$

Commented bymathmax by abdo last updated on 06/Mar/20

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

Answered by MJS last updated on 06/Mar/20

$$\int\frac{\mathrm{sin}^{\mathrm{2}} \:{x}}{\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}}{dx}= \\$$ $$\:\:\:\:\:\left[{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\rightarrow\:{dx}=\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}\right] \\$$ $$=−\mathrm{8}\int\frac{{t}^{\mathrm{2}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}\right)}{dt}= \\$$ $$=\mathrm{2}\int\frac{{t}^{\mathrm{2}} +\mathrm{2}{t}−\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{dt}−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\int\frac{{dt}}{{t}−\mathrm{1}−\sqrt{\mathrm{2}}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\int\frac{{dt}}{{t}−\mathrm{1}+\sqrt{\mathrm{2}}}= \\$$ $$=−\frac{{t}+\mathrm{1}}{{t}^{\mathrm{2}} +\mathrm{1}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{ln}\:\frac{{t}−\mathrm{1}+\sqrt{\mathrm{2}}}{{t}−\mathrm{1}−\sqrt{\mathrm{2}}}\:=... \\$$ $$=−\frac{\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{ln}\:\mid\mathrm{tan}\:\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{8}}\right)\mid\:+{C} \\$$ $$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{8}}\mathrm{ln}\:\left(\mathrm{17}+\mathrm{12}\sqrt{\mathrm{2}}\right) \\$$

Commented byKamel Kamel last updated on 06/Mar/20

$$\mathrm{17}+\mathrm{12}\sqrt{\mathrm{2}}=\mathrm{9}+\mathrm{8}+\mathrm{2}.\mathrm{3}.\mathrm{2}\sqrt{\mathrm{2}}=\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\left(\mathrm{1}+\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{4}} \\$$ $$\therefore\:{answer}\:{is}\:\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}}}{Ln}\left(\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{4}} \right)=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{Ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\$$

Answered by Kamel Kamel last updated on 06/Mar/20

$${Put}:\:\Omega=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}^{\mathrm{2}} \left({x}\right)}{{sin}\left({x}\right)+{cos}\left({x}\right)}{dx}\overset{{t}=\frac{\pi}{\mathrm{2}}−{x}} {=}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}^{\mathrm{2}} \left({t}\right)}{{cos}\left({t}\right)+{sin}\left({t}\right)}{dt} \\$$ $$\therefore\:\Omega=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{{sin}\left({x}\right)+{cos}\left({x}\right)}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{{sin}\left(\frac{\pi}{\mathrm{4}}+{x}\right)} \\$$ $$\int\frac{{dx}}{{sin}\left({x}\right)}\overset{{t}={tg}\left(\frac{{x}}{\mathrm{2}}\right)} {=}\:\:\:\:\int\frac{{dt}}{{t}}={Ln}\mid{tg}\left(\frac{{x}}{\mathrm{2}}\right)\mid+{c} \\$$ $$\therefore\:\Omega=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left({Ln}\mid{tg}\left(\frac{\pi}{\mathrm{8}}+\frac{\pi}{\mathrm{4}}\right)\mid−{Ln}\mid{tg}\left(\frac{\pi}{\mathrm{8}}\right)\mid\right) \\$$ $$\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left({Ln}\mid{tg}\left(\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{8}}\right)\mid−{Ln}\mid{tg}\left(\frac{\pi}{\mathrm{8}}\right)\mid\right. \\$$ $${cos}\left(\mathrm{2}\frac{\pi}{\mathrm{8}}\right)=\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)−\mathrm{1}\Rightarrow{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{8}}\right)=\frac{\mathrm{1}+\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{\mathrm{2}}} \\$$ $$\therefore\:{tg}\left(\frac{\pi}{\mathrm{8}}\right)=\sqrt{\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{1}+\sqrt{\mathrm{2}}}−\mathrm{1}}=\sqrt{\frac{\sqrt{\mathrm{2}}−\mathrm{1}}{\sqrt{\mathrm{2}}+\mathrm{1}}}=\sqrt{\mathrm{2}}−\mathrm{1} \\$$ $$\therefore\:\Omega=−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{Ln}\left({tg}\left(\frac{\pi}{\mathrm{8}}\right)\right)=−\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{Ln}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{Ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\$$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{KAMEL}}\:\boldsymbol{{BENAICHA}} \\$$

Answered by TANMAY PANACEA last updated on 06/Mar/20

$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}^{\mathrm{2}} {x}}{{sinx}+{cosx}}{dx} \\$$ $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}−{x}\right)}{{sin}\left(\frac{\pi}{\mathrm{2}}−{x}\right)+{cos}\left(\frac{\pi}{\mathrm{2}}−{x}\right)}{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{cos}^{\mathrm{2}} {x}}{{cosx}+{sinx}}{dx} \\$$ $$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\sqrt{\mathrm{2}}\:\left({sin}\frac{\pi}{\mathrm{4}}{cosx}+{cos}\frac{\pi}{\mathrm{4}}{sinx}\right)} \\$$ $${I}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cosec}\left({x}+\frac{\pi}{\mathrm{4}}\right){dx} \\$$ $$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mid{lntan}\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{8}}\right)\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\$$ $$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left({lntan}\left(\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{8}}\right)−{lntan}\left(\frac{\pi}{\mathrm{8}}\right)\right) \\$$ $${tan}\frac{\pi}{\mathrm{8}}={a} \\$$ $$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left({ln}\left(\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}\right)−{lna}\right)=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\left(\frac{\mathrm{1}+{a}}{{a}−{a}^{\mathrm{2}} }\right) \\$$