UNKNOWN Questions

Question Number 83226 by 09658867628 last updated on 28/Feb/20

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

Commented bymathmax by abdo last updated on 29/Feb/20

$$\left.{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{\left({x}+{cos}\alpha\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} \alpha}{changement}\:{x}+{cos}\alpha\:={sin}\alpha\right){u}\:{give} \\$$ $${I}\:=\int_{{cotan}\alpha} ^{\frac{\mathrm{1}+{cos}\alpha}{{sin}\alpha}} \:\:\frac{{sin}\alpha\:{du}}{{sin}^{\mathrm{2}} \alpha\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{{sin}\alpha}\left[{arctanu}\right]_{\frac{\mathrm{1}}{{tan}\alpha}} ^{\frac{\mathrm{1}}{{tan}\left(\frac{\alpha}{\mathrm{2}}\right)}} \\$$ $$=\frac{\mathrm{1}}{{sin}\alpha}\left\{\:{arctan}\left(\frac{\mathrm{1}}{{tan}\left(\frac{\alpha}{\mathrm{2}}\right)}\right)−{arctan}\left(\frac{\mathrm{1}}{{tan}\alpha}\right)\right\} \\$$ $${if}\:{tan}\alpha>\mathrm{0}\:{we}\:{get}\:{I}\:=\frac{\mathrm{1}}{{sin}\alpha}\left\{\frac{\pi}{\mathrm{2}}−\frac{\alpha}{\mathrm{2}}−\frac{\pi}{\mathrm{2}}\:+\alpha\right\}\:=\frac{\alpha}{\mathrm{2}{sin}\alpha}\:.... \\$$

Answered by MJS last updated on 28/Feb/20

$$\int\frac{{dx}}{{x}^{\mathrm{2}} +\mathrm{2}{x}\mathrm{cos}\:\alpha\:+\mathrm{1}}= \\$$ $$=\int\frac{{dx}}{\left({x}+\mathrm{cos}\:\alpha\right)^{\mathrm{2}} +\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\alpha}= \\$$ $$\:\:\:\:\:\left[{t}=\frac{{x}+\mathrm{cos}\:\alpha}{\sqrt{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\alpha}}\:\rightarrow\:{dx}={dt}\sqrt{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\alpha}\right] \\$$ $$\:\:\:\:\:\left[\mathrm{of}\:\mathrm{course}\:\sqrt{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \:\alpha}=\mid\mathrm{sin}\:\alpha\mid\right] \\$$ $$=\frac{\mathrm{1}}{\mid\mathrm{sin}\:\alpha\mid}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{arctan}\:{t}}{\mid\mathrm{sin}\:\alpha\mid}=\frac{\mathrm{arctan}\:\frac{{x}+\mathrm{cos}\:\alpha}{\mid\mathrm{sin}\:\alpha\mid}}{\mid\mathrm{sin}\:\alpha\mid}\:+{C} \\$$