My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum

Thanks Tree1Thanks
  • 1 Post By MarkFL
LinkBack Thread Tools Display Modes
March 2nd, 2017, 10:38 AM   #1
Joined: Mar 2017
From: Europe

Posts: 2
Thanks: 0

Constant of the Integral as a derivative?

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'
Skatinima is offline  
March 2nd, 2017, 10:47 AM   #2
Senior Member
MarkFL's Avatar
Joined: Jul 2010
From: St. Augustine, FL., U.S.A.'s oldest city

Posts: 12,184
Thanks: 481

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
MarkFL is online now  
March 2nd, 2017, 11:20 AM   #3
Senior Member
Joined: May 2016
From: USA

Posts: 1,053
Thanks: 431

$\displaystyle \int \sin(x) * \cos(x)\ dx = 0.5 * \sin^2(x) + C_1 =$

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

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.
JeffM1 is online now  
March 2nd, 2017, 11:23 AM   #4
Joined: Mar 2017
From: Europe

Posts: 2
Thanks: 0

That makes sense. Thanks for clearing this out!
Skatinima is offline  

  My Math Forum > College Math Forum > Calculus

constant, derivative, integral, trigonometry

Thread Tools
Display Modes

Similar Threads
Thread Thread Starter Forum Replies Last Post
Prove analytic function is constant, if argument is constant ag07 Complex Analysis 6 December 4th, 2015 07:05 PM
can integral be replaced by constant smslce Calculus 1 August 28th, 2014 08:25 PM
Write down, up to an arbitrary constant, the integral of x. 3uler Calculus 2 August 22nd, 2014 06:19 AM
constant value integral mhhjti Calculus 1 November 15th, 2011 09:43 AM
Evaluating improper integral for constant C TsAmE Calculus 3 August 30th, 2010 01:19 PM

Copyright © 2018 My Math Forum. All rights reserved.