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Thread: 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. $\displaystyle \displaystyle\int{\dfrac{dx}{x^2\sqrt{x^2+16}}$

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

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

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

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

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

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

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

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

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

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

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

    $\displaystyle =-\dfrac{\sqrt{x^2+16}}{16x}+C$
    Last edited by acevipa; Sep 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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