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Math Help - Integration by trigonometric substitution

  1. #1
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    Integration by trigonometric substitution

    Could someone check whether I've got the correct answer for the following integrals?

    1. \displaystyle\int{\dfrac{dx}{x^2\sqrt{x^2+16}}

    I used the trigonometric substitution x=4\tan\theta and got this as my answer

    dx=4\sec^2\theta \ d\theta

    \Rightarrow\displaystyle\int{\dfrac{dx}{x^2\sqrt{x  ^2+16}}=\int{\frac{4\sec^2\theta}{16\tan^2\theta\s  qrt{16\tan^2\theta+16}}} \ d\theta

    =\displaystyle\int{\frac{4\sec^2\theta}{16\tan^2\t  heta\cdot 4\sec\theta}} \ d\theta

    =\displaystyle\int{\frac{\sec\theta}{16\tan^2\thet  a}} \ d\theta

    =\displaystyle\frac{1}{16}\int{\frac{\cos\theta}{\  sin^2\theta}} \ d\theta

    Let u=\sin\theta\Rightarrow du=\cos\theta \ d\theta

    =\displaystyle\frac{1}{16}\int{\frac{1}{u^2}} \ du

    =\displaystyle -\frac{1}{16u}+C

    -\dfrac{1}{16\sin\theta} + C

    \tan\theta=\dfrac{x}{4}\Rightarrow \sin\theta=\dfrac{x}{\sqrt{x^2+16}}

    =-\dfrac{\sqrt{x^2+16}}{16x}+C
    Last edited by acevipa; September 5th 2010 at 10:36 PM.
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  2. #2
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    Your answer is wrong. Will you please show your calcualtions?
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  3. #3
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    Quote Originally Posted by acevipa View Post
    ...
    Perfect!
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