1. ## integration

Hi there

I was wondering if anyone would be able to help me with a problem I'm having. I've nearly finished a sum and I know the answer but I just haven't managed to quite get there. Any help would really be appreciated.

$\int_{0}^{2\pi} ( \frac{9}{8} - \frac{1}{8} \cos(4\theta)) \: d\theta \: = \: \int_{0}^{2\pi} ( \frac{9}{8}) \: d\theta - \int_{0}^{2\pi} (\frac{1}{8} \cos(4\theta)) \: d\theta$

The answer $\frac{189 \sqrt{2} \pi}{8}$ but as far as I can figure out $\: \int_{0}^{2\pi} ( \frac{9}{8}) \: d\theta$ becomes $( \frac{9}{8}) \: \theta$ but does $\int_{0}^{2\pi}(\frac{1}{8} \cos(4\theta)) \: d\theta$ become $(\frac{1}{8})(\frac{1}{4} \sin(4\theta)$ when I integrate as I've been trying to get to the answer I've been given but can't as isn't $sin(4\theta)$=0 when $\theta = 2\pi$??

If anyone could help me with the last step, I'd really appreciate it

Thanks

2. Originally Posted by michaela-donnelly
Hi there

I was wondering if anyone would be able to help me with a problem I'm having. I've nearly finished a sum and I know the answer but I just haven't managed to quite get there. Any help would really be appreciated.

$\int_{0}^{2\pi} ( \frac{9}{8} - \frac{1}{8} \cos(4\theta)) \: d\theta \: = \: \int_{0}^{2\pi} ( \frac{9}{8}) \: d\theta - \int_{0}^{2\pi} (\frac{1}{8} \cos(4\theta)) \: d\theta$

The answer $\frac{189 \sqrt{2} \pi}{8}$ but as far as I can figure out $\: \int_{0}^{2\pi} ( \frac{9}{8}) \: d\theta$ becomes $( \frac{9}{8}) \: \theta$ but does $\int_{0}^{2\pi}(\frac{1}{8} \cos(4\theta)) \: d\theta$ become $(\frac{1}{8})(\frac{1}{4} \sin(4\theta)$ when I integrate as I've been trying to get to the answer I've been given but can't as isn't $sin(4\theta)$=0 when $\theta = 2\pi$??

If anyone could help me with the last step, I'd really appreciate it

Thanks

Your calculation looks good to me. And after all, $\int_{0}^{2\pi}(\frac{1}{8} \cos(4\theta)) \: d\theta = 0$. So the answer is only $\int _0^{2\pi }\frac{9}{8}\theta~ \text{d}\theta = \boxed{~\frac{9}{4}\pi~}$

Is this whole question? I think it isn't, the answer you gave is not the result I found.