1. ## Integration Problem

I am stuck! I've worked and reworked the following integral and I'm off by a factor of 2 (or 1/2) and can't seem to figure out why. Here is my work:

$\int\theta\sin\theta\cos\theta d\theta = \frac{1}{2}\int(\theta)(2sin\theta\cos\theta) d\theta = \frac{1}{2}\int\theta\sin2\theta d\theta$ [Applying double-angle identity]

Integrating by parts: letting $u=\theta, du=d\theta, dv=sin2\theta d\theta, v=-\frac{1}{2}cos2\theta$ gives:

$-\frac{1}{2}\theta cos2\theta + \frac{1}{2}\int cos2\theta d\theta$

Integrating the right hand term:

$-\frac{1}{2}\theta cos2\theta + \frac{1}{4}sin2\theta + C = \frac{1}{4}(sin2\theta-2\theta cos2\theta) + C$

The text answer is $\frac{1}{8}(sin2\theta-2\theta cos2\theta) + C$

Where am I missing the other factor of (1/2)??

2. I am not quite sure where you are missing 1/2 but wolfram cnan give you the correct answer

integral theta sin&#x28;theta&#x29;&#x2a; &#x28;theta cos&#x28;theta&#x29; &#x29; - Wolfram|Alpha

3. You forgot to multiply by the $\frac{1}{2}$ in front of the integral.

4. Thanks, Random. You're absolutely right! Some times you can't see the forest for the trees. When I integrated using parts, I neglected the 1/2 factor in front of the original integral. Thanks!