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

[Integrate]

Ignore the du and c less bottom answer.


The top is my interpretation and work through and the bottom is the solution manual interpretation and work through.

Why don't they equate? I've been looking at it for an hour and can't spot my mistake.


Thank you

 

[LCKurtz]

You probably don't have a mistake. If you plot your two answers you will see they differ by a constant so their difference is in the constant of integration. You might try differentiating their difference and showing it is zero.

 

[Steven G]

They don't need to equate!

 

[Integrate]

They don't need to equate!

Why?

 

[Integrate]

You probably don't have a mistake. If you plot your two answers you will see they differ by a constant so their difference is in the constant of integration. You might try differentiating their difference and showing it is zero.

Meaning I misplaced a constant?

 

[LCKurtz]

Meaning I misplaced a constant?

No. Suppose you had the simpler integral [MATH]\int 2\sin x \cos x ~dx[/MATH] and got an answer of [MATH]-\cos^2(x) + C[/MATH] and someone else using a different method got [MATH]\sin^2(x) + C[/MATH]. Are either of the answers wrong?

 

[Integrate]

No. Suppose you had the simpler integral [MATH]\int 2\sin x \cos x ~dx[/MATH] and got an answer of [MATH]-\cos^2(x) + C[/MATH] and someone else using a different method got [MATH]\sin^2(x) + C[/MATH]. Are either of the answers wrong?

Just a different selection in U so yeah, you’re right.

But how can that be?

 

[Steven G]

Meaning I misplaced a constant?

That constant you missed was a major mistake
Please try in integrate (x+2)^2 in two different ways. 1st multiply out (x+2)^2 = x^2 + 4x +4 and then integrate. 2ndly integrate again using the substitution u = x+2, then convert back to x's.

Did you get the same results? Why?

 

[Steven G]

Just a different selection in U so yeah, you’re right.

But how can that be?

You NEVER get a unique answer to an indefinite inegral. You get the correct answer up to a constant!

 

[Steven G]

Just a different selection in U so yeah, you’re right.

But how can that be?

sin²x + c₁ = (1- cos²x) + c₁ = - cos²x + (c₁+1) = - cos²x + c₂.

Is it clear now?

 

[LCKurtz]

@Integrate: What you really should do with your particular problem is show the two answers are equivalent, i.e, they differ by a constant. An easy way to show [MATH]f-g = C[/MATH] is to show its derivative is zero. Try that with the difference of your two answers:
[MATH]\sin^2 x - \frac {\sin^4 x}{2}+\frac{\cos^4 x} 2[/MATH]You might like what you see and learn something in the process.

 

[Integrate]

sin²x + c₁ = (1- cos²x) + c₁ = - cos²x + (c₁+1) = - cos²x + c₂.

Is it clear now?

Yes thank you

 

[Integrate]

@Integrate: What you really should do with your particular problem is show the two answers are equivalent, i.e, they differ by a constant. An easy way to show [MATH]f-g = C[/MATH] is to show its derivative is zero. Try that with the difference of your two answers:
[MATH]\sin^2 x - \frac {\sin^4 x}{2}+\frac{\cos^4 x} 2[/MATH]You might like what you see and learn something in the process.

So I just subtract the two and take the derivative to see if it equals zero if I want to know if they are equivalent?

 

[LCKurtz]

Yes. Isn't that exactly what I said?? Show us what you get.

 

[Steven G]

So I just subtract the two and take the derivative to see if it equals zero if I want to know if they are equivalent?

Don't you see that would be the case. Suppose [math]\int f(x)dx[/math] = F(x) + c₁ and also equals G(x) + c₂. Then F(x) - G(x) = c₂ - c₁ = c. The derivative of a constant and only a constant is 0.

 

[Integrate]

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!

 

[Steven G]

OR differentiate them both and see that it equals zero. Are you sure about that part of your statement?

 

[JeffM]

Just a different selection in U so yeah, you’re right.

But how can that be?

Because the cosine and sine functions are the same function subject to a phase shift

[MATH]cos(x ) = sin\left (\dfrac{\pi}{2} + x \right )[/MATH]

 

[Steven G]

Because the cosine and sine functions are the same function subject to a phase shift

[MATH]cos(x ) = sin\left (\dfrac{\pi}{2} + x \right )[/MATH]

All functions have a phase shift.

 

[JeffM]

All functions have a phase shift.

Oh so x^{3/4} is the same as tan(x) just subject to a phase shift?