Integration Questions

Question Number 160597 by cortano last updated on 03/Dec/21

$$\:\:\:\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\frac{\mathrm{dx}}{\mathrm{2}−\mathrm{cos}\:\mathrm{x}}\:=? \\$$

Answered by Mathspace last updated on 03/Dec/21

$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{dx}}{\mathrm{2}−{cosx}}\:{we}\:{do}\:{the}\:{changement} \\$$$${tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:\Rightarrow{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{2}−\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)} \\$$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\mathrm{2}+\mathrm{2}{t}^{\mathrm{2}} −\mathrm{1}+{t}^{\mathrm{2}} }=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{\mathrm{3}{t}^{\mathrm{2}} +\mathrm{1}} \\$$$$=_{\sqrt{\mathrm{3}}{t}={u}} \:\:\mathrm{2}\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \:\:\frac{{du}}{\:\sqrt{\mathrm{3}}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)} \\$$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\left[{arctanu}\right]_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \:=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}{arctan}\left(\sqrt{\mathrm{3}}\right) \\$$

Commented by cortano last updated on 03/Dec/21

$$\mathrm{yes}.... \\$$