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Math Help - Integration by trigonometric substitution

  1. #1
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    Integration by trigonometric substitution

    I'm new to this type of question and I can only get so far into the question before I don't know what to do - I haven't been taught more but I very much would like to know.

    It asks:

    Use the substitution x = 3\sin\theta to show that

    \int_{1.5}^{3} \sqrt{9 - x^2} dx = \int_{a}^{b} k \cos^2\theta d\theta where the values of a, b and k are to be found.

    Hence evaluate \int_{1.5}^{3} \sqrt{9 - x^2} dx.

    So far I can say that x = 3\sin\theta and dx =  3\cos\theta d\theta

    \sqrt{9 - (3\sin\theta)^2} = \sqrt{9 - 9\sin^2\theta}

    = \sqrt{9 (1 - \sin^2\theta)}
    = \sqrt{9\cos^2\theta}
    = 3\cos\theta

    And this is as far as I can get.

    Could someone please help me finish the question and understand how it works from this point onwards?
    Any help is greatly appreciated
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  2. #2
    JML
    JML is offline
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    \int_{1.5}^{3} \sqrt{9 - x^2} dx = \int_{a}^{b} k \cos^2\theta d\theta where the values of a, b and k are to be found.

    As you have done we let: x = 3\sin\theta and dx =  3\cos\theta d\theta

    so

    \int_{1.5}^{3} \sqrt{9 - x^2} dx = \int_{1.5}^{3}\sqrt{9 - (3\sin\theta)^2} 3\cos\theta d\theta

    = \int_{1.5}^{3} 3\cos \theta \cdot 3\cos\theta d\theta

    = \int_{1.5}^{3} 9 \cos^2\theta d\theta
    so a=1.5 b=3 k=9

    now to evaluate this integral you can use the identity: \cos^2 x = \frac{1+cos(2x)}{2}
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