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Math Help - indefinite integral

  1. #1
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    indefinite integral

    Can anyone please help to find an antiderivative of this function and the method used?
    (Please see attatchment)

    Cheers,
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  2. #2
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    Quote Originally Posted by chrizzle View Post
    Can anyone please help to find an antiderivative of this function and the method used?
    (Please see attatchment)

    Cheers,
    Use trigonometric substitution.

    Let x = 3\tan{\theta} so that dx = 3\sec^2{\theta}\,d\theta.

    Also note that \theta = \arctan{\frac{x}{3}}.


    So the integral becomes:

    \int{\frac{1}{x^2 + 9}\,dx} = \int{\frac{1}{(3\tan{\theta})^2 + 9}\,3\sec^2{\theta}\,d\theta}

     = \int{\frac{3\sec^2{\theta}}{9\tan^2{\theta} + 9}\,d\theta}

     = \int{\frac{3\sec^2{\theta}}{9(\tan^2{\theta} + 1)}\,d\theta}

     = \int{\frac{\sec^2{\theta}}{3\sec^2{\theta}}\,d\the  ta}

     = \int{\frac{1}{3}\,d\theta}

     = \frac{1}{3}\theta + C

     = \frac{1}{3}\arctan{\frac{x}{3}} + C.
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  3. #3
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    Great, thanks

    Just 2 follow up questions:
    1. Does sec^2=tan^2 +1 and how do we know this?
    2. Is there any way to be sure what to use instead of x in trigometric substitution?
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  4. #4
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    Quote Originally Posted by chrizzle View Post
    Great, thanks

    Just 2 follow up questions:
    1. Does sec^2=tan^2 +1 and how do we know this?
    2. Is there any way to be sure what to use instead of x in trigometric substitution?
    1. Yes it does.

    You should know the Pythagorean Identity

    \sin^2{\theta} + \cos^2{\theta} = 1.


    If you divide both sides by \cos^2{\theta} you find

    \frac{\sin^2{\theta}}{\cos^2{\theta}} + \frac{\cos^2{\theta}}{\cos^2{\theta}} = \frac{1}{\cos^2{\theta}}

    \tan^2{\theta} + 1 = \sec^2{\theta}.



    2. If you prefer to use a table of integrals, then you can use the formula

    \int{\frac{1}{a^2 + x^2}\,dx} = \frac{1}{a}\arctan{\frac{x}{a}} + C.

    As you can see, this result is found using trigonometric substitution.
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  5. #5
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    actually please ignore the first question
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