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Math Help - Problem with this Integral

  1. #1
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    Problem with this Integral

    Hello,

    I have seemingly been going in circles with this problem for quite a while now. Anyway, here is the problem:

    integrate(6 dx/(x*sqrt(25x^2 -1)))

    I used trigonometric substitution:

    x = sec(theta)/5
    dx = (sec(theta)*tan(theta)*d(theta))/5

    After integrating with these values, I got 6*theta, so I did this:

    x = sec(theta)/5
    5x = sec(theta)
    arcsec(5x) = theta

    So, this gives 6*arcsec(5x) for my solution.

    However, the real solution is 6*arctan(sqrt(25x^2-1))

    If anyone could tell me where I am going wrong, it would be greatly appreciated.

    Thanks.
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  2. #2
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    Hello, machi4velli!

    Your work is absolutely correct!
    . . The book did something really stupid!


    \int\frac{6\,dx}{x\sqrt{25x^2 -1}}

    I used trigonometric substitution:
    x \:=\:\frac{}{5}\sec\theta\quad\Rightarrow\quad dx \:=\:\frac{1}{5}\sec\theta\tan\theta\,d\theta

    Substitute: . \int\frac{6\left(\frac{1}{5}\sec\theta\tan\theta\,  d\theta\right)}{\left(\frac{1}{5}\sec\theta\right)  (\tan\theta)} \;=\;6\int d\theta

    After integrating, I got: . 6\theta + C, so I did this:
    . . x = \frac{1}{5}\sec\theta\quad\Rightarrow\quad 5x = \sec\theta\quad\Rightarrow\quad \theta \:=\:\text{arcsec}(5x)

    So, this gives: . 6\text{ arcsec}(5x) + C for my solution. . Correct!

    However, the real solution is: . 6\text{ arctan}\left(\sqrt{25x^2-1}\right) + C ??

    Their answer is correct . . . but unforgivably stupid!

    You had: . \sec\theta = 5x

    Consider the right triangle:
    Code:
                            *
                         *  |
                      *     |  ________
                5x *        | √25x - 1
                *           |
             * θ            |
          * - - - - - - - - *
                    1
    Since \sec\theta = \frac{hyp}{adj}, we have: . hyp = 5x,\:adj = 1

    Using Pythagorus, we find that: . opp = \sqrt{25x^2 - 1}

    Since \tan\theta = \frac{\sqrt{25x^2-1}}{1}, it is true that \theta \:=\:\arctan(\sqrt{25x^2 - 1})
    . . but it is a truly silly way to write the answer!


    By their reasoning, all of these are also correct answers:

    . . \begin{array}{cccc}\theta &= &\arcsin\left(\frac{\sqrt{25x^2-1}}{5x}\right) \\<br />
\theta & = & \arccos\left(\frac{1}{5x}\right) \\<br />
\theta & = & \text{arccot}\left(\frac{1}{\sqrt{25x^2 - 1}}\right) \\<br />
\theta & = & \text{arccsc}\left(\frac{5x}{\sqrt{25x^2-1}}\right)<br />
\end{array}

    but who (in their right minds) would insist on using them?

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  3. #3
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    Ah! I did not even think about this. No wonder I was becoming so confused, thank you for the help. Thanks for pointing out that I forgot to add the constant also.

    By the way, how do you format the notation correctly? Does the text somehow generate the picture? Very interesting...
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Soroban View Post

    By their reasoning, all of these are also correct answers:

    . . \begin{array}{cccc}\theta &= &\arcsin\left(\frac{\sqrt{25x^2-1}}{5x}\right) \\<br />
\theta & = & \arccos\left(\frac{1}{5x}\right) \\<br />
\theta & = & \text{arccot}\left(\frac{1}{\sqrt{25x^2 - 1}}\right) \\<br />
\theta & = & \text{arccsc}\left(\frac{5x}{\sqrt{25x^2-1}}\right)<br />
\end{array}

    but who (in their right minds) would insist on using them?

    These aren't necessarily bad or "silly" forms for the answer. That depends on the application of the results. It may be that the arctan solution is actually the "simplest" form for the solution depending on what problem generates the integral.

    -Dan
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  5. #5
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    Hello, topsquark!


    I agree that the application may dictate the form of the answer.


    But I maintain that, for a textbook Calculus problem,

    . . writing the answer as: \text{arccot}\left(\frac{\sqrt{1-x^2}}{x}\right) instead of \arcsin x

    . . is unnecessary, misleading, and highly annoying.

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