Question Number 190754 by universe last updated on 10/Apr/23
$$\:\:\:\:\:\:{I}\:\:\:=\:\:\:\:\int_{\mathrm{0}} ^{\:\pi} {e}^{{a}\mathrm{cos}\:{t}} \:\mathrm{cos}\:\left({a}\mathrm{sin}\:\:{t}\right){dt} \\ $$
Answered by namphamduc last updated on 11/Apr/23
$${I}=\int_{\mathrm{0}} ^{\pi} {e}^{{a}\mathrm{cos}\left({t}\right)} \mathrm{cos}\left({a}\mathrm{sin}\left({t}\right)\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\Re\int_{−\pi} ^{\pi} {e}^{{a}\mathrm{cos}\left({t}\right)} {e}^{{ia}\mathrm{sin}\left({t}\right)} {dt}=\frac{\mathrm{1}}{\mathrm{2}}\Re\int_{−\pi} ^{\pi} {e}^{{a}\mathrm{cos}\left({t}\right)+{ia}\mathrm{sin}\left({t}\right)} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\Re\int_{−\pi} ^{\pi} {e}^{{ae}^{{it}} } {dt},{z}={e}^{{it}} \Rightarrow{dz}={ie}^{{it}} {dt}\Rightarrow{dt}=−{i}\frac{{dz}}{{z}} \\ $$$$\Rightarrow{I}=\frac{\mathrm{1}}{\mathrm{2}}\Re\left(−{i}\int_{\mid{z}\mid=\mathrm{1}} \frac{{e}^{{az}} }{{z}}{dz}\right)=\frac{\mathrm{1}}{\mathrm{2}}\Re\left(−{i}.\mathrm{2}\pi{i}.\mathrm{Res}\left(\frac{{e}^{{az}} }{{z}},{z}=\mathrm{0}\right)\right)=\pi \\ $$