# Trig integral question.

1. Apr 7, 2009

### AzMaphysics

1. The problem statement, all variables and given/known data

$$\int$$sin32x dx

2. The attempt at a solution

$$\int$$sin32x dx =
$$\int$$(1-cos22x)sin2x dx =
(-1/2)$$\int$$(1-cos22x)-2sin2x dx =
(-1/2)$$\int$$(1-u2)du where u= cos2x
(-1/2)(u-(u3/3)) =
(-cos 2x/2)+(cos32x/6) + C

Book says (cos3x/3)-(cosx/2) + C
Where did I go wrong?

Last edited: Apr 7, 2009
2. Apr 7, 2009

### Staff: Mentor

It looks like you have the right idea, but you should split the integral into two separate integrals after this step:
$\int (1-cos^22x)sin2x dx$

3. Apr 7, 2009

### AzMaphysics

Yea integrating by parts seems to be the preferred way to attack these types of problems. The chapter was focusing on u-substitution though so I tried going with that. I just can't tell if I made a mistake or if the book did (there's been a couple times in the past) or if the book's answer is just more simplified and I'm not seeing how.

4. Apr 7, 2009

### n!kofeyn

Well I actually checked your work instead of assuming you made a mistake, and you made no mistakes (I got the same thing). The way to do this is definitely a u-substitution and you don't want to split it up as mentioned. Your answer is correct.

To save work, I also computed the derivative of our answer with Maple and it gave me the integrand that we started with. I did the same for the book's "answer", and it did not. So it looks like the book is incorrect on this one.