# Thread: calculating a integral -need help

1. ## calculating a integral -need help

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

Thank you very much.

2. Hi turkeywilliam,

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.

3. oops! sorry, it should be the integral - not Double integral.

4. $\frac{1}{4} \int_{0}^{\frac{\pi}{3}} 16 \cos^{4} \theta -1 \ d \theta = 4 \int_{0}^{\frac{\pi}{3}} \cos^{3} \theta \ d \theta - \frac{1}{4} \int_{0}^{\frac{\pi}{3}} \ d \theta$

5. $\frac{1}{4} \int_{0}^{\frac{\pi}{3}} 16 \cos^{4} \theta -1 \ d \theta = 4 \int_{0}^{\frac{\pi}{3}} \cos^{4} \theta \ d \theta - \frac{1}{4} \int_{0}^{\frac{\pi}{3}} \ d \theta$

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

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

So we get: $\frac{\pi}{3} + \sin \frac{2 \pi}{3} + \frac{\pi}{6} + \frac{1}{8} \sin \frac{4 \pi}{3} - \frac{\pi}{12} = \frac{\pi}{3} + \frac{\sqrt{3}}{2} + \frac{\pi}{6} - \frac{\sqrt{3}}{16} - \frac{\pi}{12} = \boxed{\frac{5 \pi}{12} + \frac{7 \sqrt{3}}{16}}$

6. 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?

7. yes I got that.

8. yes, me too. Thank you very much.