Question Number 46854 by maxmathsup by imad last updated on 01/Nov/18
$${find}\:\:=\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{sinx}}{\mathrm{2}+{cos}\left(\mathrm{2}{x}\right)}{dx} \\ $$
Commented by maxmathsup by imad last updated on 01/Nov/18
$${et}\:{A}\:=\int_{\mathrm{0}} ^{\pi} \:\frac{{sinx}}{\mathrm{2}\:+{cos}\left(\mathrm{2}{x}\right)}{dx}\:\Rightarrow\:{A}\:=\int_{\mathrm{0}} ^{\pi} \:\:\frac{{sinx}}{\mathrm{2}\:+\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{sinx}}{\mathrm{1}+\mathrm{2}{cos}^{\mathrm{2}} {x}}\:{dx}=_{\sqrt{\mathrm{2}}{cosx}={t}} \:\:\:\int_{\sqrt{\mathrm{2}}} ^{−\sqrt{\mathrm{2}}} \:\:\:\frac{−\mathrm{1}}{\:\sqrt{\mathrm{2}}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\:\int_{−\sqrt{\mathrm{2}}} ^{\sqrt{\mathrm{2}}} \:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\sqrt{\mathrm{2}}{arctan}\left(\sqrt{\mathrm{2}}\right)\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Nov/18
$$\int_{\mathrm{0}} ^{\pi} \frac{{sinx}}{\mathrm{2}+\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1}}{dx} \\ $$$$=−\mathrm{1}×\int_{\mathrm{0}} ^{\pi} \frac{{d}\left({cosx}\right)}{\mathrm{1}+\mathrm{2}{cos}^{\mathrm{2}} {x}} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}}×\int_{\mathrm{0}} ^{\pi} \frac{{d}\left({cosx}\right)}{\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:}\right)^{\mathrm{2}} +{cos}^{\mathrm{2}} {x}} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}}×\sqrt{\mathrm{2}}\:\mid{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}}\:{cosx}}{\mathrm{1}}\right)\mid_{\mathrm{0}} ^{\pi} \\ $$$$=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{2}}}×\left\{{tan}^{−\mathrm{1}} \left(−\sqrt{\mathrm{2}}\left[\right)−{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\:\right)\right\}\right. \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}×\mathrm{2}{tan}^{−\mathrm{1}} \left(\sqrt{\mathrm{2}}\:\right) \\ $$
Commented by maxmathsup by imad last updated on 01/Nov/18
$${sir}\:{Tanmay}\:{your}\:{answer}\:{is}\:{correct}\:{thanks}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 02/Nov/18
$${mostwelcome}… \\ $$