Integration Questions

Question Number 120297 by bramlexs22 last updated on 30/Oct/20

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

Answered by Dwaipayan Shikari last updated on 30/Oct/20

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{x}}{{sinx}+{cosx}}{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\frac{\pi}{\mathrm{2}}−{x}}{{sinx}+{cosx}}{dx}={I} \\$$$$\mathrm{2}{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\frac{\pi}{\mathrm{2}}}{{sinx}+{cosx}}{dx} \\$$$$\frac{\mathrm{4}}{\pi}{I}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{t}+\mathrm{1}−{t}^{\mathrm{2}} }{dt}\:\:\:\:\:\:\:\:\:\:{tan}\frac{{x}}{\mathrm{2}}={t} \\$$$$\frac{\mathrm{2}}{\pi}{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}−\left({t}−\mathrm{1}\right)^{\mathrm{2}} } \\$$$$\frac{\mathrm{2}}{\pi}{I}=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left[\left({log}\left(\frac{{t}−\mathrm{1}−\sqrt{\mathrm{2}}}{{t}−\mathrm{1}+\sqrt{\mathrm{2}}}\right)\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\$$$$\frac{\mathrm{2}}{\pi}{I}=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left(\mathrm{0}−{log}\left(\frac{\sqrt{\mathrm{2}}+\mathrm{1}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}\right)\right. \\$$$${I}=\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}{log}\left(\frac{\sqrt{\mathrm{2}}+\mathrm{1}}{\:\sqrt{\mathrm{2}}−\mathrm{1}}\right)=\frac{\pi}{\:\mathrm{2}\sqrt{\mathrm{2}}}{log}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right) \\$$

Commented by peter frank last updated on 30/Oct/20

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{both} \\$$

Answered by Bird last updated on 30/Oct/20

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

Answered by TANMAY PANACEA last updated on 30/Oct/20

$${or} \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\left(\frac{\pi}{\mathrm{2}}−{x}\right){dx}}{{sinx}+{cosx}} \\$$$$\mathrm{2}{I}=\frac{\pi}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\:\sqrt{\mathrm{2}}\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{sinx}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{cosx}\right)} \\$$$${I}=\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cosec}\left({x}+\frac{\pi}{\mathrm{4}}\right){dx} \\$$$${I}=\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\mid{lntan}\left(\frac{{x}}{\mathrm{2}\:}+\frac{\pi}{\mathrm{8}}\right)\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\$$$$=\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\left({lntan}\left(\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{8}}\right)−{lntan}\left(\frac{\pi}{\mathrm{8}}\right)\right) \\$$$$=\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\left({lntan}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)−{ln}\left(\frac{\pi}{\mathrm{8}}\right)\right) \\$$