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 March 2nd, 2017, 10:38 AM #1 Newbie   Joined: Mar 2017 From: Europe Posts: 2 Thanks: 0 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' March 2nd, 2017, 10:47 AM #2 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs 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. Thanks from Maschke March 2nd, 2017, 11:20 AM #3 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 550 $\displaystyle \int \sin(x) * \cos(x)\ dx = 0.5 * \sin^2(x) + C_1 =$ $0.5\{1 - \cos^2(x)\} + C_1 = -\ 0.5 * \cos^2(x) + 0.5 + C_1 = -\ 0.5 * \cos^2(x) + C_2.$ 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. March 2nd, 2017, 11:23 AM #4 Newbie   Joined: Mar 2017 From: Europe Posts: 2 Thanks: 0 That makes sense. Thanks for clearing this out! Tags constant, derivative, integral, trigonometry Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post ag07 Complex Analysis 6 December 4th, 2015 07:05 PM smslce Calculus 1 August 28th, 2014 08:25 PM 3uler Calculus 2 August 22nd, 2014 06:19 AM mhhjti Calculus 1 November 15th, 2011 09:43 AM TsAmE Calculus 3 August 30th, 2010 01:19 PM

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