What is the integral of a function: (e^(-x))cos(x/2) dx ??

Question

The answer which I got was  (e^(-x))sin(x/2)-cos(x/2)(e^(-x))  

 

However all the math solvers on the internet are giving different answers. Could you clear things up and tell me a definite answer. I've done many many manipulations to this equation, but the best answer I came up with was this one. 

Thank you. 

Answer 1

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)