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Math Help - calculating a integral -need help

  1. #1
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    calculating a integral -need help

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

    Thank you very much.
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  2. #2
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    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.
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  3. #3
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    oops! sorry, it should be the integral - not Double integral.
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  4. #4
    Senior Member tukeywilliams's Avatar
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     \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
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  5. #5
    Senior Member tukeywilliams's Avatar
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     \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}}
    Last edited by tukeywilliams; July 31st 2007 at 11:46 PM.
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  6. #6
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    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?
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  7. #7
    Senior Member tukeywilliams's Avatar
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    yes I got that.
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  8. #8
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    yes, me too. Thank you very much.
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