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Question Number 190754 by universe last updated on 10/Apr/23

$$I={\int }_0^{\pi }{e}^{a\cos t}\cos \left(a\sin t\right)dt$$

Answered by namphamduc last updated on 11/Apr/23

$$I={\int }_0^{\pi }{e}^{a\cos \left(t\right)}\cos \left(a\sin \left(t\right)\right)dt$$$$=\frac{1}{2}\mathrm{\Re }{\int }_{-\pi }^{\pi }{e}^{a\cos \left(t\right)}{e}^{ia\sin \left(t\right)}dt=\frac{1}{2}\mathrm{\Re }{\int }_{-\pi }^{\pi }{e}^{a\cos \left(t\right)+ia\sin \left(t\right)}dt$$$$=\frac{1}{2}\mathrm{\Re }{\int }_{-\pi }^{\pi }{e}^{{ae}^{it}}dt,z=e^{it}\Rightarrow dz={ie}^{it}dt\Rightarrow dt=-i\frac{dz}{z}$$$$\Rightarrow I=\frac{1}{2}\mathrm{\Re }(-i{\int }_{\mid z\mid =1}\frac{e^{az}}{z}dz)=\frac{1}{2}\mathrm{\Re }(-i.2\pi i.\mathrm{Res}(\frac{e^{az}}{z},z=0))=\pi$$

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