Second method of solving this integral needed

In summary, the indefinite integral can be solved using two separate cases: when b = 0 and when b ≠ 0. If you want to solve for x when b = 0, you need to integrate by parts and u-substitute. If you want to solve for x when b ≠ 0, you need to re-define the domain as b ≠ 0.
  • #1
flyinjoe
17
0

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!
 
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  • #2
flyinjoe said:
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?
 
  • #3
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!
 
  • #4
flyinjoe said:
should I re-integrate with b = 0, and get 1/2 x^2 + C,

Yes.
 
  • #5
flyinjoe said:
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)
 

FAQ: Second method of solving this integral needed

What is the second method of solving an integral?

The second method of solving an integral is known as the substitution method. This involves replacing a variable in the integral with a new variable in order to simplify the integral and make it easier to solve.

When should I use the second method of solving an integral?

The substitution method is useful for solving integrals that involve complicated functions or expressions. It can also be used to solve integrals with trigonometric functions or exponential functions.

How do I choose the substitution variable?

The substitution variable should be chosen so that when it is substituted into the integral, it simplifies the expression and makes it easier to integrate. A good rule of thumb is to choose a variable that appears in the integral and has a derivative that also appears in the integral.

What are the steps for solving an integral using the substitution method?

The steps for solving an integral using the substitution method are as follows:

  1. Choose an appropriate substitution variable.
  2. Substitute the variable into the integral, including the differential (dx).
  3. Simplify the new integral using algebraic manipulation.
  4. Integrate the simplified integral.
  5. Substitute the original variable back into the solution.

Can the substitution method always be used to solve integrals?

No, the substitution method may not always be the most efficient or effective method for solving an integral. It is important to consider other methods such as integration by parts or partial fractions when attempting to solve an integral.

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