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How do you integrate #int e^(sinpix)cospix# from #[0,pi/2]#?

Answer 1

Evelyn Agard

#1/pi(e^(sin(pi^2/2))-1)#

Remembering that

#d/(dx)e^(sin(pix))=pi cos(pix)e^(sin(pix))# we have

#int cos(pix)e^(sin(pix))dx=1/pi(e^(sin(pi^2/2))-1)#

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Answer 2

Isaiah Allen

To integrate ( \int e^{\sin(\pi x)}\cos(\pi x) ) from ( x = 0 ) to ( x = \frac{\pi}{2} ), you can use integration by parts twice, applying the method to ( u = e^{\sin(\pi x)} ) and ( dv = \cos(\pi x) , dx ). The result will be ( \frac{e^{\sin(\pi x)}(\sin(\pi x) - \cos(\pi x))}{\pi+1} ) evaluated from ( x = 0 ) to ( x = \frac{\pi}{2} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.