# Help with an Integration problem

1. Dec 4, 2005

### asdf1

(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?

2. Dec 4, 2005

### siddharth

Why should the "correct answer" be (1/2a)sin^ax? How do you know the "correct answer" is actually right?

3. Dec 4, 2005

### benorin

It's okay

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:

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

where C is the constant of integration.

however, one may instead apply the substitution

$$u=\sin(ax) \Rightarrow du=a\cos(ax)dx$$

to the given integral like this

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

But how could that be? because

$$-\frac{1}{4a} \cos(2ax) =C+\frac{1}{2a} \sin^{2}(ax)$$

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

4. Dec 4, 2005

### asdf1

thank you very much!!! :)