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    {\bf p}(t) = (f(t)\cos\,t,f(t)\sin\,t)

    {\bf p}'(t) = (f'(t)\cos\,t - f(t)\sin\,t , f'(t)\sin\,t + f(t) \cos\,t)
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    (Original post by TheEd)


    Parametrising the curve C_1 as follows:

    {\bf p}(t) = (1-t)(1,4) + t(3,3) = (1+2t,4-t)

    {\bf p}'(t) = (2,-1)

    So \displaystyle \int_{C_1} = \bigg\{ \sqrt{\frac{y}{x}} \frac{dx}{dt} + \sqrt{\frac{x}{y}} \frac{dy}{dt} \bigg\} \;dt = \int^1_0 \bigg\{ 2\sqrt{\frac{4-t}{1+2t}} - \sqrt{\frac{1+2t}{4-t}} \bigg\} \;dt

    How do I do this integration?

    Apparently the integrand is a derivative of a function of the form \sqrt{f(t)} where f(t) is a 2nd order polynomial.
    For the first root use the
    \displaystyle \sqrt{\frac{4-t}{1+2t}}=u
    substitution
    You will get a fraction which maybe integrated with another substitution and/or breaking into part fractions
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    add 1 to the power and divide by the power
 
 
 
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