[Trig Integrals] Why Don't These Integrals Equate?

[LCKurtz]

Thank you both of you.


What I've learned is that indefinite integrals don't need to equate when taken different approaches to integrate them but they will be only different by a constant. So in order to check that they are equal up to a constant it is best to take their difference OR differentiate them both and see that it equals zero. Thank you!

See that what equals zero? I think it's too early to be thanking me since I don't think you understand or can calculate what I suggested. In post #11 I suggested you take the difference of your two answers, which is:

[MATH]\sin^2 x - \frac {\sin^4 x}{2}+\frac{\cos^4 x} 2[/MATH]
and differentiate it and show the result is zero. You haven't done that. Show me the steps like it was a quiz.

 

[Steven G]

No, but f(x) = x^(3/4) = f(x-1) = (x-1)^(3/4).

In any integral you can make the u substitution u=x-1 for example.

 

[JeffM]

No, but f(x) = x^(3/4) = f(x-1) = (x-1)^(3/4).

In any integral you can make the u substitution u=x-1 for example.

Jomo I think you missed my point. I was saying that two functions were the same except for a phase shift. The question that I was answering was how can -cos^2(x) + unknown constant be equivalent to sin^2(x) + unknown constant.

And I certainly do not think it is true that x^(3/4) = (x - 1)^(3/4). Let's try that with x = 1. The equation resolves to 1 = 0.

Sorry if my post was unclear.

 

[Steven G]

Jomo I think you missed my point. I was saying that two functions were the same except for a phase shift. The question that I was answering was how can -cos^2(x) + unknown constant be equivalent to sin^2(x) + unknown constant.

And I certainly do not think it is true that x^(3/4) = (x - 1)^(3/4). Let's try that with x = 1. The equation resolves to 1 = 0.

Sorry if my post was unclear.

opps, x^(3/4) = (x - 1)^(3/4) is not true.

I guess I understood your post but I am just pointing out that there does not have to be a known phase shift for this to happen. Just integrate (x+2)^2 two different ways and you'll see my point.

 

[JeffM]

I suspect we are talking past each other. I was simply explaining why expressions in terms of the sine can always be expressed in terms of the cosine. Expressions in terms of the sine cannot always be expressed in a finite expression in (x - 1). I was talking about a relationship between two specific functions.

 

[Integrate]

See that what equals zero? I think it's too early to be thanking me since I don't think you understand or can calculate what I suggested. In post #11 I suggested you take the difference of your two answers, which is:

[MATH]\sin^2 x - \frac {\sin^4 x}{2}+\frac{\cos^4 x} 2[/MATH]
and differentiate it and show the result is zero. You haven't done that. Show me the steps like it was a quiz.

Is f-g not the second half of my page?

 

[LCKurtz]

I suggested you differentiate [MATH]\sin^2 x - \frac {\sin^4 x}{2}+\frac{\cos^4 x} 2[/MATH] and show the result gives [MATH]0[/MATH]. You do that by writing something like:

[MATH]f(x) = \sin^2 x - \frac {\sin^4 x}{2}+\frac{\cos^4 x} 2[/MATH]so [MATH]f'(x) = \dots[/MATH]
Differentiate it, simplify it and see that you get [MATH]0[/MATH]. I'm not your boss and you don't have to do it if you don't want to. I think it would be instructive if you would do it, and if you can't or won't, I'm just wasting time in this thread.