I've been wondering about the following integral recently:
I=∫√1+cosxdx
The way I integrated it is I used the identity √1+cosx=√2cosx2, and so
I=2√2sinx2+C
the problem is that √1+cosx is actually integrable over the entire real line, but the derivative of 2√2sinx2 is only equal to √1+cosx in certain intervals. This is because the actual identity is √1+cosx=√2|cosx2|. Now, I wasn't exactly sure how to integrate the absolute value, so I thought I would "fix" the function after integration.
The first fix we can do is making sure that the sign of the result of integration is correct:
I=2√2sinx2sgncosx2+C
The problem with just sinx2 was that sometimes it was "flipped", on certain intervals it was actually the antiderivative of the −√1+cosx (this is because we dropped the absolute value signs).
There is one further problem with this, however. The above function is not continuous, meaning it's not continuously differentiable with derivative equal to √1+cosx everywhere. Namely, it is discontinuous at x=(4n±1)π where n∈Z.
I noticed, however, that the limit of the derivative on either side of x=(4n±1)π existed and was equal to each other. Hence, I figured that I could somehow "stitch" these continuous sections end to end and get a continuous result whose derivative was √1+cosx everywhere. The resulting function I got was
I=2√2sinx2sgncosx2+4√2⌊12πx+12⌋+C
Now, I was wondering, is there any way to arrive at this result just by integrating √1+cosx using the usual techniques? The method I used can be boiled down to "integrating" then "fixing", but I'm just wondering if you can arrive at a result that is continuous and differentiable on the entire real line by doing just the "integrating" part.
Any help would be appreciated. Thanks!
Edit: To be clear, I'm not looking for exactly the function above, but rather simply any function that is "nice", continuous and differentiable on R, and has derivative equal to √1+cosx everywhere, and which is attainable through "simple" integration methods.
Answer
The function to be integrated is
√1+cosx=√2|cosx2|
so we can as well consider the simpler
∫|cost|dt
or, with t=u+π/2, the even simpler
∫|sinu|du
One antiderivative is
∫u0|sinv|dv
Note that the function has period π, so let u=kπ+u0, with 0≤u0<π. Then
∫u0|sinv|dv=∫kπ0|sinv|dv+∫kπ+u0kπ|sinv|dv=2k+∫u00sinvdv=2k+1−cos(u−kπ)
Now do the back substitution; write k=umodπ, if you prefer.
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