Help with an Integration problem

  • Thread starter Thread starter asdf1
  • Start date Start date
  • Tags Tags
    Integration
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 3K views
asdf1
Messages
734
Reaction score
0
(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?
 
Physics news on Phys.org
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! :)