Second method of solving this integral needed

[flyinjoe]

Homework Statement

Evaluate the indefinite integral using two separate cases:
∫xcos(bx) dx

Homework Equations

Integration by parts:
∫f(x)g'(x) dx = f(x)g(x) - ∫f'(x)g(x)

The Attempt at a Solution

I solved the problem using integration by parts and u-substitution and got the answer:
(cos(bx) + bx(sin(bx))) / b^2 + C

But my professor says we should discuss the problem in two separate ways. What might be the other method of solving this problem?

Thank you!

 

[pasmith]

But my professor says we should discuss the problem in two separate ways. What might be the other method of solving this problem?

No, he says there are two cases to consider. Does your answer work for the case b = 0?

 

[flyinjoe]

My answer definitely does not work when b = 0. Wonderful!

Now, should I re-integrate with b = 0, and get 1/2 x^2 + C, or re-define the domain as b≠0 ?

Sorry if these are silly questions,
Thank you!

 

[haruspex]

should I re-integrate with b = 0, and get 1/2 x^2 + C,

Yes.

 

[Mute]

My answer definitely does not work when b = 0. Wonderful!

Now, should I re-integrate with b = 0, and get 1/2 x^2 + C, or re-define the domain as b≠0 ?

Sorry if these are silly questions,
Thank you!

Note also that you write the indefinite integral as

$$\int dx~x\cos(bx) = \frac{bx\sin(bx) + \cos(bx) - 1}{b^2} + C'.$$
You can do because the factor I introduced, $-1/b^2$, is just a constant which could be re-absorbed into C'.

If you now take the limit as $b \rightarrow 0$, you will find that your answer reduces to $x^2/2 + C'$.

(You don't need to split the factor of $-1/b^2$ off from the constant $C = C'-1/b^2$ from the original post, but if you don't you will have to throw away a "constant infinite factor" when taking the $b \rightarrow 0$ limit, and that can feel like you're being kind of shady. =P)