Integration (different substitution = different answer)

In summary, the conversation discusses how certain trig functions can be integrated using Integration by Substitution without the use of trig identities. The example of sin(x)cos(x) dx is used, and it is shown that integrating with substitution u = cos(x) works fine, but u = sin(x) does not. It is explained that this is because the derivative of cos(x) is negative sin(x), so the natural thing to do is to multiply everything inside the integral by -1. However, it is pointed out that both integrals differ by a constant, and that the constant of integration is necessary in indefinite integrals. The conversation also mentions the importance of considering the constant when evaluating definite integrals.
  • #1
n0_3sc
243
1
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?
 
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  • #2
Why does the substitution u=sin(x) not work?
 
  • #3
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
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
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
Sorry i meant u = cos(x) works fine but u = sin(x) doesn't. (My bad).

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.

Even if you didn't know about the different substitutions how would you know that you need to multiply by -1 to get the right answer (assuming you have no way to check your 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).

I agree it may have been a bad example but this was a spur of the moment quesiton...in this integrand f(x) does have a derivative present that's why substitution is best here...this doesn't explain why the one substitution works.

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.

There is a constant of integration for everything you integrate, I am just concerned with the integrands solution. Are you saying that some substitutions don't work because they require extra info in getting the constant of integrations?

I'm not trying to be arrogant, I just think there is a logical solution to it like everything else in math :smile:
 
  • #7
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
Even if you didn't know about the different substitutions how would you know that you need to multiply by -1 to get the right answer (assuming you have no way to check your answer).

No I was talking about since the derivative of cos x is NEGATIVE sin x, but we only have a sin x in the integrand, the natural thing to do is to multiply everything inside the integral by -1, and the outside by -1, to even it out.
 
  • #9
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
n0_3sc said:
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?

What must you always include when doing an indefinite integral? An arbitrary constant of integration, and this is because the derivative of a constant function is zero. If you were doing a definite integral either substitution would give the exact same final answer because that constant ends up disappearing when it is added and subtracted.

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]=[2-1]+[3-3]=1
similarly
Y(2)-Y(1)=[2+1]-[1+1]=[2-1]+[1-1]=1

And they evaluate to the same answer.
 
  • #11
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
n0_3sc said:
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?

This seems to mostly be a case of you not understanding the result.

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
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
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.
 
Last edited:

1. What is integration and why is it important?

Integration is a mathematical technique used to find the area under a curve or the sum of infinitely many small quantities. It is important because it has numerous applications in physics, engineering, economics, and many other fields.

2. What is substitution in integration?

Substitution is a technique used to simplify an integrand by replacing a variable with another variable or expression. This is done to make the integral easier to solve.

3. Why do different substitutions lead to different answers in integration?

Different substitutions can lead to different answers in integration because the choice of substitution can greatly affect the complexity of the resulting integral. A simpler substitution may lead to a more straightforward solution, while a more complicated substitution may lead to a more complex solution.

4. How do I know which substitution to use in integration?

Choosing the appropriate substitution in integration can be done by looking for patterns in the integrand, such as a power rule or trigonometric functions. It also takes practice and familiarity with different substitution techniques.

5. Are there any tips for solving integrals with different substitutions?

One tip for solving integrals with different substitutions is to always check your answer by differentiating it. This helps to ensure that the substitution was done correctly and that the resulting answer is accurate. Additionally, it is helpful to practice and become familiar with different substitution techniques to improve problem-solving skills.

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