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Math Help - Simple question - rather embarrassing!

  1. #1
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    Simple question - rather embarrassing!

    Hi guys,

    Found this site through google, looks great

    Anyway, I've just finished my maths degree and am training to be a teacher at the moment. My plan is to teach up to further maths at A-level (UK), so as such I'm going over a lot of the stuff I haven't really looked at since my first undergraduate year. I'm rather embarrassingly stumped on a question, and I know the answer is staring me right in the face, if you guys could point me in the right direction it would be great, thanks

    Write (\sec\theta+\tan\theta)(\sec\theta-\tan\theta) in it's simplest form. Use it to evaluate

    \int \frac{1}{(\sec\theta + \tan\theta)^2}d\theta

    The first part evaluates to 1, which means I can write

    \int \frac{1}{(\sec\theta + \tan\theta)^2}d\theta=\int\frac{\sec\theta - \tan\theta}{\sec\theta + \tan\theta}d\theta

    but then I find I'm a bit stuck, the next step would be muchly appreciated

    Kind Regards

    Stonehambey
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  2. #2
    Member Glaysher's Avatar
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    Newton-le-Willows
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    Quote Originally Posted by Stonehambey View Post
    Hi guys,

    Found this site through google, looks great

    Anyway, I've just finished my maths degree and am training to be a teacher at the moment. My plan is to teach up to further maths at A-level (UK), so as such I'm going over a lot of the stuff I haven't really looked at since my first undergraduate year. I'm rather embarrassingly stumped on a question, and I know the answer is staring me right in the face, if you guys could point me in the right direction it would be great, thanks

    Write (\sec\theta+\tan\theta)(\sec\theta-\tan\theta) in it's simplest form. Use it to evaluate

    \int \frac{1}{(\sec\theta + \tan\theta)^2}d\theta

    The first part evaluates to 1, which means I can write

    \int \frac{1}{(\sec\theta + \tan\theta)^2}d\theta=\int\frac{\sec\theta - \tan\theta}{\sec\theta + \tan\theta}d\theta

    but then I find I'm a bit stuck, the next step would be muchly appreciated

    Kind Regards

    Stonehambey
    Multiply top and bottom by {\sec\theta - \tan\theta} again

    The denominator becomes 1. The to[p becomes something you can integrate after multiplying out and using identities
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  3. #3
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    Of course! Thanks!
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