Help with an Integration problem

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Homework Help Overview

The discussion revolves around an integration problem involving the integral of the product of sine and cosine functions, specifically ∫sin(ax)cos(ax)dx. Participants are exploring the correct evaluation of this integral and the discrepancies between their calculations and a provided answer.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to evaluate the integral using a trigonometric identity and expresses confusion over the difference between their result and the expected answer. Other participants question the validity of the "correct answer" and explore alternative methods, including substitution.

Discussion Status

Participants are actively engaging with the problem, raising questions about the correctness of the expected answer and discussing different approaches to the integral. There is a recognition of various interpretations and methods being considered, but no consensus has been reached regarding the resolution of the discrepancies.

Contextual Notes

Some participants reference external sources, such as textbooks or software, to validate answers, indicating a reliance on authoritative materials for correctness. The discussion also hints at potential misunderstandings regarding the application of trigonometric identities and integration techniques.

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(integration) (sinax)(cosax)dx = (1/2)(integration)[sin2ax)dx
= (-1/4a)cos2ax = (-1/4a)(1-2(sinax)^2)

but the correct answer should be (1/2a)sin^ax

does anybody know what went wrong?
 
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Why should the "correct answer" be (1/2a)sin^ax? How do you know the "correct answer" is actually right?
 
It's okay

asdf1 said:
(integration) (sinax)(cosax)dx = (1/2)(integration)[sin2ax)dx
= (-1/4a)cos2ax = (-1/4a)(1-2(sinax)^2)
but the correct answer should be (1/2a)sin^ax
does anybody know what went wrong?

Sure, you're right, and the back of the book (or Maple or whatever) is also right [assuming you meant (1/2a)sin(ax)^2:

[tex]\int \sin(ax)\cos(ax)dx = \frac{1}{2}\int \sin(2ax)dx = -\frac{1}{4a} \cos(2ax) +C ,[/tex]

where C is the constant of integration.

however, one may instead apply the substitution

[tex]u=\sin(ax) \Rightarrow du=a\cos(ax)dx[/tex]

to the given integral like this

[tex]\int \sin(ax)\cos(ax)dx =\frac{1}{a} \int u du = \frac{1}{2a} u^2 +C = \frac{1}{2a} \sin^{2}(ax) +C[/tex]

But how could that be? because

[tex]-\frac{1}{4a} \cos(2ax) =C+\frac{1}{2a} \sin^{2}(ax)[/tex]

is the half-angle identity from you used [quote: (-1/4a)cos2ax = (-1/4a)(1-2(sinax)^2)] for the proper value of C.
 
thank you very much! :)
 

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