UNKNOWN Questions

Question Number 64139 by Chi Mes Try last updated on 14/Jul/19

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

Commented by Prithwish sen last updated on 14/Jul/19

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{dx}}{\left(\mathrm{x}+\mathrm{cos}\alpha\right)^{\mathrm{2}} +\left(\mathrm{sin}\alpha\right)^{\mathrm{2}} } \\$$$$=\left[\frac{\mathrm{1}}{\left(\mathrm{sin}\alpha\right)}.\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{x}+\mathrm{cos}}{\mathrm{sin}\alpha}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\$$

Commented by mathmax by abdo last updated on 14/Jul/19

$${let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{2}{xcos}\alpha\:+\mathrm{1}}\:\Rightarrow{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{2}{xcos}\alpha\:+{cos}^{\mathrm{2}} \alpha\:+{sin}^{\mathrm{2}} \alpha} \\$$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\left({x}+{cos}\alpha\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} \alpha}\:\:{changement}\:{x}+{cos}\alpha\:={sin}\alpha\:{t}\:{give} \\$$$${I}\:=\:\int_{{cotan}\alpha} ^{\frac{\mathrm{1}+{cos}\alpha}{{sin}\alpha}} \:\:\:\frac{{sin}\alpha\:{dt}}{{sin}^{\mathrm{2}} \alpha\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{{sin}\alpha}\left[{arctant}\right]_{\frac{\mathrm{1}}{{tan}\alpha}} ^{\frac{\mathrm{1}+{cos}\alpha}{{sin}\alpha}} \\$$$$=\frac{\mathrm{1}}{{sin}\alpha}\left\{\:{arctan}\left(\frac{\mathrm{1}+{cos}\alpha}{{sin}\alpha}\right)−{arctan}\left(\frac{\mathrm{1}}{{tan}\alpha}\right){but}\right. \\$$$$\frac{\mathrm{1}+{cos}\alpha}{{sin}\alpha}\:=\frac{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\alpha}{\mathrm{2}}\right)}{\mathrm{2}{sin}\left(\frac{\alpha}{\mathrm{2}}\right){cos}\left(\frac{\alpha}{\mathrm{2}}\right)}\:=\frac{\mathrm{1}}{{tan}\left(\frac{\alpha}{\mathrm{2}}\right)}\:\Rightarrow \\$$$${arctan}\left(\frac{\mathrm{1}+{cos}\alpha}{{sin}\alpha}\right)=\overset{−} {+}\frac{\pi}{\mathrm{2}}\:−\frac{\alpha}{\mathrm{2}} \\$$$${arctan}\left(\frac{\mathrm{1}}{{tan}\alpha}\right)=\overset{−} {+}\frac{\pi}{\mathrm{2}}\:−\alpha\:\:\Rightarrow\:{I}\:=\frac{\mathrm{1}}{{sin}\alpha}\left\{\overset{−} {+}\frac{\pi}{\mathrm{2}}\:\overset{−} {+}\frac{\pi}{\mathrm{2}}\:+\frac{\alpha}{\mathrm{2}}\right\} \\$$$${so}\:{there}\:{is}\:{a}\:{error}\:{in}\:{the}\:{question}\:{if}\:{not}\:{a}\:{equation}... \\$$

Answered by Hope last updated on 15/Jul/19

$${D}_{{r}} =\left({x}+{cos}\alpha\right)^{\mathrm{2}} +{sin}^{\mathrm{2}} \alpha \\$$$${t}={x}+{cos}\alpha \\$$$$\int_{{cos}\alpha} ^{\mathrm{1}+{cos}\alpha} \frac{{dt}}{{t}^{\mathrm{2}} +{sin}^{\mathrm{2}} \alpha} \\$$$$\frac{\mathrm{1}}{{sin}\alpha}\mid{tan}^{−\mathrm{1}} \left(\frac{{t}}{{sin}\alpha}\right)\mid_{{cos}\alpha} ^{\mathrm{1}+{cos}\alpha} \\$$$$=\frac{\mathrm{1}}{{sin}\alpha}\left[{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}+{cos}\alpha}{{sin}\alpha}\right)−{tan}^{−\mathrm{1}} \left(\frac{{cos}\alpha}{{sin}\alpha}\right)\right] \\$$$$=\frac{\mathrm{1}}{{sin}\alpha}\left[{tan}^{−\mathrm{1}} \left({cot}\frac{\alpha}{\mathrm{2}}\right)−{tan}^{−\mathrm{1}} \left({cot}\alpha\right)\right] \\$$$$=\frac{\mathrm{1}}{{sin}\alpha}\left[{tan}^{−\mathrm{1}} \left({tan}\left(\frac{\pi}{\mathrm{2}}−\frac{\alpha}{\mathrm{2}}\right)\right)−{tan}^{−\mathrm{1}} \left(\frac{\pi}{\mathrm{2}}−\alpha\right)\right] \\$$$$=\frac{\mathrm{1}}{{sin}\alpha}\left[\frac{\pi}{\mathrm{2}}−\frac{\alpha}{\mathrm{2}}−\frac{\pi}{\mathrm{2}}+\alpha\right] \\$$$$=\frac{\alpha}{\mathrm{2}{sin}\alpha} \\$$