Integration by parts question help

[Suraj M]

Homework Statement

While integrating by parts( by the formula) why don't we consider the contant of integration for every integral in the equation.

Homework Equations

$$∫uv = u∫v - ∫ ∫v . d/dx(u) $$

The Attempt at a Solution

[/B]
example.
$$∫x \sin(x) dx = ?? $$
this is can be done like this
$$ x ∫ \sin(x) - ∫ ∫ \sin(x) .1 $$
$$ ⇒ -x \cos(x) + \sin(x) + c$$
why not like this..??
$$ ⇒ x( - \cos(x) +c₁ ) +\sin(x) +c₂$$
in this case the two constants don't add up to give one constant, it give s a new ##c_1x## term

 

[SammyS]

Homework Statement

While integrating by parts( by the formula) why don't we consider the constant of integration for every integral in the equation.

Homework Equations

$$∫uv = u∫v - ∫ ∫v . d/dx(u) $$

The Attempt at a Solution

[/B]
example.
$$∫x \sin(x) dx = ?? $$
this is can be done like this
$$ x ∫ \sin(x) - ∫ ∫ \sin(x) .1 $$
$$ ⇒ -x \cos(x) + \sin(x) + c$$
why not like this..??
$$ ⇒ x( - \cos(x) +c₁ ) +\sin(x) +c₂$$
in this case the two constants don't add up to give one constant, it give s a new ##c_1x## term

You have lots of ∫ symbols with no trailing differential, like dx or du or dv ...
That makes it difficult or impossible to untangle much of what you have written.

To see that the result for your example is wrong, differentiate that result.

 

[andrewkirk]

You have to put the same integration constant in both places where it appears on the RHS, so it cancels out. In your example this works as so:
$$\int x \sin x\, dx = x \int \sin x\, dx - \int 1\cdot \int \sin u\, du\, dx
= x [- \cos x+c_1] - \int 1\cdot [- \cos x+c_1]\, dx
= -x \cos x-c_1x +c_1x + \sin x+c_2
= -x \cos x + \sin x+c_2
$$
So we end up with only one integration constant.

 

[Suraj M]

You have lots of ∫ symbols with no trailing differential, like dx or du or dv ...
That makes it difficult or impossible to untangle much of what you have written.

Im very sorry about that! i can't edit it now. But i hope you get what I'm saying. Also if you could verify Andrew's reasoning, it would be very helpful.

 

[Suraj M]

You have to put the same integration constant in both places where it appears on the RHS, so it cancels out.

Why should we consider the same constant?? in both cases, because differentiating with different constants we still get ##\sin(x)##.

 

[RUber]

It is the same function. You start with ##\int x \sin(x) dx## and then say that your sine is function 'v' in your formula. So, yes there are different constants that would work, but if you use different constants, you have two different functions. v = v all the time.
'

 

[Suraj M]

ut if you use different constants, you have two different functions. v = v all the time.

Yes, but there is a possibility for ## ∫ v dv ≠ ∫v dv ## ?

 

[RUber]

Nope. There is a possibility that ##\int v'_1 dv \neq \int v'_2 dv## where v1 and v2 have the same derivative. But your assumption from the outset is that v = v.

 

[Suraj M]

So are you saying that ##∫vdv =∫vdv##?always?

 

[RUber]

Within the context of a given problem, yes.
When you apply integration by parts to the form ##\int f(x) dx##, you are assuming that f(x)dx is of the form uv'. Thus you are assuming that there is one v that gives you that v' and you want to solve for it. Clearly, there are an infinite number of options, which is why you denote it with a C, but the underlying assumption is that there is only one function v.

 

[andrewkirk]

Why should we consider the same constant?? in both cases, because differentiating with different constants we still get ##\sin(x)##.

The reason is in the proof that integration by parts works. THe proof is based on the product rule
##(fg)'=f'g+fg'##
which is rearranged to give
##fg'=(fg)'-f'g##
We then integrate both sides to get
##\int fg' = fg +c_1 - \int f'g##
We can ignore ##c_1## because it will end up being added to the constant of integration we get when we integrate ## \int f'g##.
So we can just write
##\int fg' = fg - \int (f'g)##.

This is the correct form of the rule for integration by parts. Note that ##g## occurs twice on the RHS, and it is the same function both times. It cannot differ by a constant under this formula.

The version of the rule you wrote is not quite correct, specifically because it introduces an additional constant. Texts sometimes do this, but it is sloppy, and can introduce errors, of the type you have identified here. If we want to write it without a ##g'##, we should write:
##\int fh = fg - \int (f'g)## where ##g## is a function whose derivative is ##h##.

 

[SteamKing]

So are you saying that ##∫vdv =∫vdv##?always?

If we got a different anti-derivative every time we integrated something, calculus would be meaningless. Mathematics would be totally and capriciously random.

 

[Suraj M]

Thank you Andrew, cleared all my doubts :-)

 

[epenguin]

There shouldn't be any big mystery about integrating by parts. It is easily understood and remembered thinking of definite integrals and areas The curve in my as usual crude sketch represents a function u(v) of variable v. The total rectangle whose corners are (0, 0) and (b, u(b)) has area b.u(b) . The total shaded are is [uv]_a^b or (b.u(b) - a.u(a)). The |||| hatched area is ∫_a^b u dv and this obviously equals the difference between the said total hatched [uv]_a^b and the ≡≡ hatched
which is obviously ∫_u(a)^u(b)v du . Simples.

 

[Suraj M]

but then this(the question i have mentioned) is not a definite integral,
im sorry if i might have misinterpreted what you were saying.

 

[epenguin]

but then this(the question i have mentioned) is not a definite integral,
im sorry if i might have misinterpreted what you were saying.

Well think about it more flexibly or maybe someone else can explain what I mean. Every indefinite integral is essentially a definite integral by the time you've included the constant of integration and used it for anything, I'd say. Consider the fact you can make the integration limits anything you like (or rather anything he function permits), Whatever the a and b in my fig.- don't have to be the ones I chose.