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Integration only giving one answer?

  1. Jul 17, 2017 #1
    1. The problem statement, all variables and given/known data
    If ##a>0## and ##b≠0##, solve the following stating the maximal domain for which the solution is valid:
    ##\sqrt{a^2-x^2}\cdot \frac{dy}{dx}+b=0,\ y\left(0\right)=0##

    2. Relevant equations

    ##\int _{ }^{ }\frac{1}{\sqrt{a^2-x^2}}dx=\arcsin \left(\frac{x}{a}\right)+c,\ a>0##

    ##\int _{ }^{ }\frac{-1}{\sqrt{a^2-x^2}}dx=\arccos \left(\frac{x}{a}\right)+c,\ a>0##

    3. The attempt at a solution
    I have no problems doing the actual integration and arrive at the correct solution of ##y=-b\arcsin \left(\frac{x}{a}\right),\ \left|x\right|<a\ ##. However, I also get ##y=b\arccos \left(\frac{x}{a}\right)-b,\ \left|x\right|<a\ ## as being a valid solution, whereas both the answers and my calculator don't include this solution. Is there a reason as to why this second answer would be omitted?
     
  2. jcsd
  3. Jul 17, 2017 #2

    Mark44

    Staff: Mentor

    ##\arcsin(x) + \arccos(x) = \frac \pi 2##, meaning the two functions differ by a constant, ##\pi/2##. Your two answers are different only by a constant.
     
  4. Jul 17, 2017 #3

    scottdave

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    When you integrate, you must find the constant (c) using the initial conditions. With the correct constants, either arcos or arcsin should work. It looks like you need a constant added
     
  5. Jul 17, 2017 #4
    Ah thank you, I wasn't aware of that relationship. So basically it's similar to saying that ##\frac{d}{dx}\left(\tan \left(x\right)\right)=\sec ^2\left(x\right)## or ##\frac{d}{dx}\left(\tan \left(x\right)\right)=\tan ^2\left(x\right)\ +1##.
     
  6. Jul 17, 2017 #5

    Ray Vickson

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    How about a third solution? Maple gets
    $$y = -b \arctan \left(\frac{x}{\sqrt{a^2-x^2}} \right) $$
     
  7. Jul 18, 2017 #6

    scottdave

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    This could work, too. If you think of a right triangle, then arcsin, arccos, and arctan are all returning angles, given a ratio of two sides of the triangle.
    So if x represents the adjacent, and a represents the hypotenuse, then arccos(x/a) will return the angle theta, for cos(theta) = x/a
    Similarly, x could represent the opposite angle of a triangle, when using arcsin.
    So for arctan, you need the opposite/adjacent. Since hypotenuse2 = adjacent2 + opposite2, try working out to see how you could get the ratio for a tangent, using x representing the opposite, and a representing the hypotenuse.
     
  8. Jul 18, 2017 #7

    Ray Vickson

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    Yes, of course, it is completely elementary. However, I hoped the OP could puzzle it out.
     
  9. Jul 18, 2017 #8

    scottdave

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    You may find this video about trig integration, interesting. It may shed some light for you.
     
  10. Jul 18, 2017 #9

    scottdave

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    Whoops, I didn't look closely. I thought the original person had posted that.
     
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