the integral of 16 cos^4 theta -1 d(theta).

Thank you very much.

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- Jul 31st 2007, 09:34 PMkittycatcalculating a integral -need help
the integral of 16 cos^4 theta -1 d(theta).

Thank you very much. - Jul 31st 2007, 09:55 PMkittycat
Hi turkeywilliam,

Thank you very much for your reply.

Let me give you the exact question with the range.

evaluate 1/4 times the double integral ( from 0 to pi/3) of 16 cos^4 theta -1 d(theta). Could you please show me the final answer with detail steps . Thanking you in advance. - Jul 31st 2007, 10:05 PMkittycat
oops! sorry, it should be the integral - not Double integral.

- Jul 31st 2007, 10:12 PMtukeywilliams
$\displaystyle \frac{1}{4} \int_{0}^{\frac{\pi}{3}} 16 \cos^{4} \theta -1 \ d \theta = 4 \int_{0}^{\frac{\pi}{3}} \cos^{3} \theta \ d \theta - \frac{1}{4} \int_{0}^{\frac{\pi}{3}} \ d \theta $

- Jul 31st 2007, 10:25 PMtukeywilliams
$\displaystyle \frac{1}{4} \int_{0}^{\frac{\pi}{3}} 16 \cos^{4} \theta -1 \ d \theta = 4 \int_{0}^{\frac{\pi}{3}} \cos^{4} \theta \ d \theta - \frac{1}{4} \int_{0}^{\frac{\pi}{3}} \ d \theta $

So $\displaystyle 4 \int_{0}^{\frac{\pi}{3}} \cos^{4} \theta \ d \theta = 4 \int_{0}^{\frac{\pi}{3}} \left(\frac{1+ \cos 2 \theta}{2} \right)^{2} = \int_{0}^{\frac{\pi}{3}} 1 + 2 \cos 2 \theta + \frac{1}{2}(1 + \cos 4 \theta) \ d \theta $ $\displaystyle = \theta + \sin 2 \theta + \frac{\theta}{2} + \frac{1}{8} \sin 4 \theta \ d \theta $

And so we have $\displaystyle \theta + \sin 2 \theta + \frac{\theta}{2} + \frac{1}{8} \sin 4 \theta - \frac{1}{4} \theta $ evaluated from $\displaystyle 0 $ to $\displaystyle \frac{\pi}{3} $.

So we get: $\displaystyle \frac{\pi}{3} + \sin \frac{2 \pi}{3} + \frac{\pi}{6} + \frac{1}{8} \sin \frac{4 \pi}{3} - \frac{\pi}{12} = \frac{\pi}{3} + \frac{\sqrt{3}}{2} + \frac{\pi}{6} - \frac{\sqrt{3}}{16} - \frac{\pi}{12} = \boxed{\frac{5 \pi}{12} + \frac{7 \sqrt{3}}{16}} $ - Jul 31st 2007, 10:29 PMkittycat
Thank you very much turkeywilliam. I'll go through your work in detail. I have the answer of this question is 7sqrt(3)/16 + 5pi/12. Do you get this answer as well?

- Jul 31st 2007, 10:40 PMtukeywilliams
yes I got that.

- Jul 31st 2007, 10:49 PMkittycat
yes, me too. Thank you very much. :)