How do you integrate $\int e^{sin π x} cos π x$ $int e^(sinpix)cospix$ from $[0, \frac{π}{2}]$ $[0,pi/2]$ ?

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Answer 1 · 2016-12-31T11:45:40

$\frac{1}{π} (e^{sin (\frac{π^{2}}{2})} - 1)$ $1/pi(e^(sin(pi^2/2))-1)$

Remembering that

$\frac{d}{d x} e^{sin (π x)} = π cos (π x) e^{sin (π x)}$ $d/(dx)e^(sin(pix))=pi cos(pix)e^(sin(pix))$ we have

$\int cos (π x) e^{sin (π x)} d x = \frac{1}{π} (e^{sin (\frac{π^{2}}{2})} - 1)$ $int cos(pix)e^(sin(pix))dx=1/pi(e^(sin(pi^2/2))-1)$

How do you integrate ∫esinπxcosπxint e^(sinpix)cospix from [0,π2][0,pi/2]?