the integral of 16 cos^4 theta -1 d(theta).
Thank you very much.
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.
$\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}} $