Ira S. answered • 05/12/17

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This is an integration by parts problem.

u = cos(x/2) dv = e^-x dx

du = -1/2sin(x/2) v = -e^-x

so we get

-e^-x cos(x/2) - ∫1/2e^-xsin(x/2)dx

u = sin(x/2) dv= 1/2 e^-x dx

du = 1/2cos(x/2)dx v = -1/2 e^-x

So

∫e^-x cos(x/2) = -e^-x cos(x/2) - [-1/2e^-x sin(x/2) - ∫-1/4e^-x cox(x/2) dx

∫e^-xcos(x/2) = -e^-x cos(x/2 + 1/2e^-x sin(x/2) - 1/4∫e^-xcos(x/2) dx we can now add the last integral to both sides to get

5/4∫ e^-xcos(x/2) = -e^-x cos(x/2) + 1/2 e^-x sin(x/2)

so

∫ e^-x cos(x/2) = -4/5 e^-x cos(x/2) + 2/5 e^-x sin (x/2)

I checked it and it worked out correct.....so I hope this helped.

I think the mistake that you may have been making is in the chain rule for the derivative or intefral of cos(x/2)

The derivative is -sin(x/2) * 1/2 and you would need to use a u substitution to get the integral as 2sin(x/2)