Integration Questions

Question Number 154437 by ArielVyny last updated on 18/Sep/21

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {arcos}\left(\frac{{cosx}}{\mathrm{1}+\mathrm{2}{cosx}}\right){dx} \\$$

Answered by phanphuoc last updated on 18/Sep/21

$${put}\:{x}={tan}\left({t}/\mathrm{2}\right) \\$$$$\\$$

Answered by Kamel last updated on 18/Sep/21

$$\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{24}} \\$$

Answered by puissant last updated on 18/Sep/21

$${Q}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {arccos}\left(\frac{{cosx}}{\mathrm{1}+\mathrm{2}{cosx}}\right){dx}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {arccos}\left(\sqrt{\frac{\mathrm{1}+\frac{{cosx}}{\mathrm{1}+\mathrm{2}{cosx}}}{\mathrm{2}}}\right){dx} \\$$$$=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {arctan}\left(\sqrt{\frac{\mathrm{1}+{cosx}}{\mathrm{1}+\mathrm{3}{cosx}}}\right){dx}=\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {arctan}\left(\sqrt{\frac{\mathrm{1}+{cos}\mathrm{2}{u}}{\mathrm{1}+\mathrm{3}{cos}\mathrm{2}{u}}}\right){du} \\$$$${Diver}:\:''\mathrm{1}+{cos}\mathrm{2}{u}=\mathrm{2}{cosu}\:{et}\:{cos}\mathrm{2}{u}=\mathrm{1}−{sin}^{\mathrm{2}} {u}\:\left({Trivial}\right)'' \\$$$$\Rightarrow\:{Q}=\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {arctan}\left(\sqrt{\frac{{cos}^{\mathrm{2}} {u}}{\mathrm{1}−{sin}^{\mathrm{2}} {u}}}\right){du}=\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {arctan}\left(\frac{{cosu}}{\:\sqrt{\mathrm{1}−\mathrm{2}{sinu}}}\right){du} \\$$$$=\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\sqrt{\mathrm{2}−\mathrm{3}{sin}^{\mathrm{2}} {u}}\:{cosu}}{\left({x}^{\mathrm{2}} +\mathrm{2}\right)−{sin}^{\mathrm{2}} {u}\left({x}^{\mathrm{2}} +\mathrm{3}\right)}{dxdu} \\$$$$\sqrt{\mathrm{3}}{sinu}=\sqrt{\mathrm{2}}{sin}\theta\:\rightarrow\:{cosudu}=\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{3}}}{cos}\theta{d}\theta \\$$$$\Rightarrow\:{Q}=\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{3}}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}\sqrt{\mathrm{3}}\:{cos}^{\mathrm{2}} \theta}{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}−\mathrm{2}\left(\mathrm{1}−{cos}^{\mathrm{2}} \theta\right)\left({x}^{\mathrm{2}} +\mathrm{3}\right)}{dxd}\theta \\$$$$\overset{{t}={tanu}} {=}\:\mathrm{8}\sqrt{\mathrm{3}}\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left({t}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}\right)}{dxdt} \\$$$$=\mathrm{8}\sqrt{\mathrm{3}}\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{−\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{6}}}{\mathrm{1}+{t}^{\mathrm{2}} }+\frac{\frac{{x}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{6}}}{{t}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{6}}\right){dxdt}=\mathrm{8}\sqrt{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{6}}\left(\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }−\int_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{3}+\frac{\mathrm{6}}{{x}^{\mathrm{2}} }}\right){dx} \\$$$$=\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{9}}−\mathrm{4}\left[{arctan}\left(\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}}\right){arctan}\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{arctan}\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}\right)}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{dx} \\$$$$=\mathrm{4}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{arctan}\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}\right)}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{dx}\:=\:\mathrm{4}×\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{96}}\:=\:\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{24}}.. \\$$$$\\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\because\therefore\:\:\:\:\:{Q}\:=\:\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{24}}...\: \\$$$$\\$$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:......................\mathscr{L}{e}\:{puissant}...................... \\$$

Commented by peter frank last updated on 18/Sep/21

$$\mathrm{good} \\$$

Commented by ArielVyny last updated on 19/Sep/21

$${thank}\:{sir} \\$$