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March 2nd, 2017, 10:38 AM   #1
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Constant of the Integral as a derivative?

Hey,
I was solving this integral on a website:

∫ sin(x)cos(x)dx

And I got two answers. One of them was this one:

½sin^(2)x + c

and the other one was this one:

-½cos^(2)x + c

But on the site it shows this solution:

-½cos^(2)x + c'

where the constant is a derivative.

I plugged the integral on the integral calculator website and the constant is not a derivative.

Does anybody know the thought process behind writing the constant as a derivative?


P.S: In case you're wondering, here are the steps for the second solution (copy-paste):

Now substitute into the formula:
∫ uvdx = u ∫vdx − ∫u'(∫vdx) dx
This gives:
∫ sin(x)cos(x)dx = cos(x) × -cos(x) − ∫ -sin(x) × -cos(x)dx
= -cos2x − ∫ sin(x)cos(x)dx

Again we have the same integral on both sides (except one is subtracted) ...
... so bring the right hand one over to the left and we get:
2∫ sin(x)cos(x)dx = -cos2x ⇒ ∫ sin(x)cos(x)dx = -½cos2x + c'
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March 2nd, 2017, 10:47 AM   #2
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My guess would be that the prime notation being used on the constant of integration does not denote differentiation, but rather distinguishes it from the original constant after dividing through by 2.
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March 2nd, 2017, 11:20 AM   #3
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$\displaystyle \int \sin(x) * \cos(x)\ dx = 0.5 * \sin^2(x) + C_1 =$

$0.5\{1 - \cos^2(x)\} + C_1 = -\ 0.5 * \cos^2(x) + 0.5 + C_1 = -\ 0.5 * \cos^2(x) + C_2.$

Or this may be what was intended. The constant is not the same when the integral is expressed in terms of the sine as it is when the integral is expressed in terms of the cosine. Using the prime sign is sloppy notation.

Last edited by skipjack; March 2nd, 2017 at 04:38 PM.
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March 2nd, 2017, 11:23 AM   #4
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That makes sense. Thanks for clearing this out!
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