# calculating a integral -need help

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

Thank you very much.
• Jul 31st 2007, 09:55 PM
kittycat
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 PM
kittycat
oops! sorry, it should be the integral - not Double integral.
• Jul 31st 2007, 10:12 PM
tukeywilliams
$\displaystyle \frac{1}{4} \int_{0}^{\frac{\pi}{3}} 16 \cos^{4} \theta -1 \ d \theta = 4 \int_{0}^{\frac{\pi}{3}} \cos^{3} \theta \ d \theta - \frac{1}{4} \int_{0}^{\frac{\pi}{3}} \ d \theta$
• Jul 31st 2007, 10:25 PM
tukeywilliams
$\displaystyle \frac{1}{4} \int_{0}^{\frac{\pi}{3}} 16 \cos^{4} \theta -1 \ d \theta = 4 \int_{0}^{\frac{\pi}{3}} \cos^{4} \theta \ d \theta - \frac{1}{4} \int_{0}^{\frac{\pi}{3}} \ d \theta$

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 PM
kittycat
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 PM
tukeywilliams
yes I got that.
• Jul 31st 2007, 10:49 PM
kittycat
yes, me too. Thank you very much. :)