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Math Help - How do I solve this integral?

  1. #1
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    How do I solve this integral?

    Hi there. Well, the problem statement says: Calculate the arc length of the cardioid \rho=a(1+\cos\theta).

    So I used the formula for the arc length in the polar form:

    \int_a^b \! \sqrt{(\rho)^2+(\displaystyle\frac{d\rho}{d\theta} )^2} \, d\theta

    Then I get
    \int_0^{2\pi} \! \sqrt{(a+a\cos\theta)^2+(-a\sin\theta)^2} \, d\theta

    =\int_0^{2\pi} \! \sqrt{a^2+2a^2\cos\theta+a^2\cos^2\theta+a^2\sin^2 \theta} \, d\theta

    =\int_0^{2\pi} \! \sqrt{a^2(2\cos\theta+\cos^2\theta+\sin^2\theta)} \, d\theta

    =\int_0^{2\pi} \! |a|\sqrt{2\cos\theta+\cos^2\theta+\sin^2\theta} \, d\theta=|a|\int_0^{2\pi} \!\sqrt{2\cos\theta+1} \, d\theta

    I can't solve this integral, I don't know how to. Any help?
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  2. #2
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    Quote Originally Posted by Ulysses View Post
    Hi there. Well, the problem statement says: Calculate the arc length of the cardioid \rho=a(1+\cos\theta).

    So I used the formula for the arc length in the polar form:

    \int_a^b \! \sqrt{(\rho)^2+(\displaystyle\frac{d\rho}{d\theta} )^2} \, d\theta

    Then I get
    \int_0^{2\pi} \! \sqrt{(a+a\cos\theta)^2+(-a\sin\theta)^2} \, d\theta

    =\int_0^{2\pi} \! \sqrt{a^2+2a^2\cos\theta+a^2\cos^2\theta+a^2\sin^2 \theta} \, d\theta

    =\int_0^{2\pi} \! \sqrt{a^2(1+2\cos\theta+\cos^2\theta+\sin^2\theta)  } \, d\theta ... algebra error here (fixed)
    ...
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  3. #3
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    Thanks skeeter.
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