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.