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Integration (different substitution = different answer) 
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#1
Mar2407, 10:08 PM

P: 265

Out of curiosity there are several trig functions that can be integrated (WITHOUT the use of trig identities) using Integration by Substitution.
One particular example is this: sin(x)cos(x) dx Integrating this with substitution u = cos(x) works out fine. HOWEVER integrating with substitution u = sin(x) does not. Can someone please explain why this is, and how to distinguish what substitution to make? 


#2
Mar2407, 10:18 PM

P: 1,075

Why does the substitution u=sin(x) not work?



#3
Mar2407, 10:55 PM

HW Helper
P: 3,348

In fact u=sin x actually works BETTER than u=cos x, no need to multiply by factors of negative 1 to get things to work.



#4
Mar2407, 11:04 PM

HW Helper
P: 860

Integration (different substitution = different answer)
your chosen example is a bad one,... but I can imagine that there are cases where only a "correct" substitution will greatly simplify things.. which is your objective in the first place. remember, integrate by substitution/change of variables works best when the integrand has both f(x) and its derviative present (as a product).



#5
Mar2407, 11:53 PM

P: 367

You get different answers until you remember that there's a constant of integration. The answers differ by a constant (1/2), so they aren't really different.



#6
Mar2507, 12:23 AM

P: 265

Sorry i meant u = cos(x) works fine but u = sin(x) doesn't. (My bad).
I'm not trying to be arrogant, I just think there is a logical solution to it like everything else in math 


#7
Mar2507, 12:35 AM

P: 367

Let [tex]u = sin(x)[/tex].
Then [tex]du = cos(x)dx[/tex], so [tex]\int \sin(x)\cos(x)dx = \int udu = \frac{1}{2}u^2 + C = \frac{1}{2}\sin^2(x) + C[/tex] Now instead, let [tex]u = cos(x)[/tex]. Then [tex]du = sin(x)dx[/tex], so [tex]\int \sin(x)\cos(x)dx = \int udu = \frac{1}{2}u^2 + C = \frac{1}{2}\cos^2(x) + C[/tex] But [tex]\frac{1}{2}\sin^2(x) = \frac{1}{2}\cos^2(x) + \frac{1}{2}[/tex] so they differ by a constant. Where is there a problem? 


#8
Mar2507, 12:45 AM

HW Helper
P: 3,348




#9
Mar2607, 12:44 AM

P: 265

Yes they do differ by a half.
Thus, the two integrals are indeed different. My point is if you were doing an analytical problem, then your end result will differ (by a constant) depending on the substitution. So which result is right? Using MATLAB the integral of sin(x)cos(x) = 0.5[cos(x)]^2. So, the substitution u = sin(x) is incorrect. Hence certain substitutions matter? Why? 


#10
Mar2607, 01:02 AM

P: 1,075

As a trivial example consider the function y=1, then both Y=x+1, and Z=x+3 are antiderivatives of y. Now suppose we want to evaluate this integral over the interval (1,2) so Z(2)Z(1)=[2+3][1+3]=[21]+[33]=1 similarly Y(2)Y(1)=[2+1][1+1]=[21]+[11]=1 And they evaluate to the same answer. 


#11
Mar2607, 03:43 AM

P: 265

I agree with you.
Would it be fair to say that for an indefinite integral you may remove any sum/difference constants and that would be the solution? (regardless of the substitution). 


#12
Mar2607, 05:02 AM

P: 277

The best way to think of it is that the MATLAB result is wrong, or rather incomplete (they are just taking the constant to be 0). It should say: [tex]\int \sin(x)\cos(x)dx = \frac{1}{2}\cos^2(x) + C[/tex] Which is exactly the same as: [tex]\int \sin(x)\cos(x)dx = \frac{1}{2}\sin^2(x) + C_2[/tex] Because: [tex]\frac{1}{2}\cos^2(x) + C = \frac{1}{2}\sin^2(x) + C_2[/tex] You can always add an arbitrary constant to an indefinite integral. If you are worried about keeping track of constants, just "eat up" any new constants by creating a new one (in the example above one of the C's ate up the other C and a constant factor of 1/2). You need to understand the difference between a definite and an indefinite integral. 


#13
Mar2707, 08:16 PM

P: 291

You'll also see that when you do a definite integral, again, you'll get the same value. (try it with the example above.)



#14
Mar2707, 09:58 PM

P: 15

Instead of doing it with a single example, you can prove that for all integrable functions, the constant of integration doesn't matter for definite integrals.
[tex]\int f(x)\,dx = F(x) + C[/tex] implies that [tex]\int^b_a f(x)\,dx = ( F(b) + C)  ( F(a) + C ) = F(b)  F(a) + C  C = F(b)  F(a)[/tex]. This means that both [tex]F(x)[/tex] and [tex]F(x) + C [/tex] give the same answer. Also, another quick way to see that both of your answers are right is just to differentiate them both and notice that they're equal. 


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