Integration only giving one answer?

  • Thread starter Saracen Rue
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    Integration
In summary, the third solution that Maple offers is cosine of the angle between the x-axis and the line segment from the origin to the point (a,0).
  • #1
Saracen Rue
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Homework Statement


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##

Homework Equations


[/B]
##\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##

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?
 
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  • #2
Saracen Rue said:

Homework Statement


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##

Homework Equations


[/B]
##\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##

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?
##\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.
 
  • #3
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
 
  • #4
Mark44 said:
##\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.
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##.
 
  • #5
Saracen Rue said:

Homework Statement


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##

Homework Equations


[/B]
##\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##

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?

How about a third solution? Maple gets
$$y = -b \arctan \left(\frac{x}{\sqrt{a^2-x^2}} \right) $$
 
  • #6
Ray Vickson said:
How about a third solution? Maple gets
$$y = -b \arctan \left(\frac{x}{\sqrt{a^2-x^2}} \right) $$

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.
 
  • #7
scottdave said:
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.

Yes, of course, it is completely elementary. However, I hoped the OP could puzzle it out.
 
  • #8
Ray Vickson said:
How about a third solution? Maple gets
$$y = -b \arctan \left(\frac{x}{\sqrt{a^2-x^2}} \right) $$
You may find this video about trig integration, interesting. It may shed some light for you.
 
  • #9
Ray Vickson said:
Yes, of course, it is completely elementary. However, I hoped the OP could puzzle it out.
Whoops, I didn't look closely. I thought the original person had posted that.
 

What is integration and why does it only give one answer?

Integration is a mathematical technique used to find the area under a curve by dividing it into smaller, simpler shapes. It only gives one answer because it is based on the fundamental theorem of calculus, which states that the definite integral of a function represents the net change of the function over the given interval. This means that there can only be one answer for the area under a curve.

Why does integration sometimes give negative answers?

Integration can give negative answers when the function being integrated has values below the x-axis. This means that the area under the curve is below the x-axis and is therefore considered negative. It is important to pay attention to the limits of integration and the orientation of the curve to understand why integration may give a negative answer.

Can integration give a decimal or fraction as an answer?

Yes, integration can give a decimal or fraction as an answer. This is because the area under a curve can have any value, including non-integer values. The answer will depend on the function being integrated and the limits of integration.

Why does integration sometimes give an infinite answer?

Integration can give an infinite answer when the function being integrated has an infinite area under the curve. This can happen when the curve extends to infinity or when there are vertical asymptotes in the function. In these cases, the area under the curve cannot be calculated and the integral will give an infinite answer.

Can integration give different answers for the same function?

No, integration will always give the same answer for the same function and limits of integration. This is because the fundamental theorem of calculus states that the definite integral of a function is unique and only depends on the function and the interval of integration. Different methods of integration may give slightly different answers due to rounding errors, but the overall result will be the same.

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