What is the correct integration of arcsin(x) using integration by parts?

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In summary, the conversation discusses the solution to the equation \int \arcsin{(x)}dx using integration by parts. The participant has found a discrepancy between their solution and the answer given in their textbook and on a website. They also discuss the convention for mapping the interval [-1,1] onto the range of arcsine and how it affects the sign in the anti-derivative. The conversation concludes with a question regarding the integration of the last part and the use of a calculator and textbook in computing a definite integral.
  • #1
Charismaztex
45
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Hello all,

I've solved this equation to get:

[tex]
\int \arcsin{(x)}dx = \sqrt{1-x^2}+ x\arcsin{(x}) + C
[/tex]

using integration by parts. I have found, however, that my textbook has
the part

[tex]-\sqrt{1-x^2}[/tex]

instead of a +, leaving the answer

[tex]
\int \arcsin{(x)}dx = -\sqrt{1-x^2}+ x\arcsin{(x}) + C
[/tex]

This minus sign has been confirmed by a website I came across
Code:
[PLAIN]http://math2.org/math/integrals/tableof.htm
[/PLAIN] [Broken]
Elsewhere on this forum I've seen the answer with the plus sign:
Code:
https://www.physicsforums.com/showthread.php?t=89216
giving rise to this
inconsistency.

Any help is appreciated and thanks in advance,
Charismaztex
 
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  • #2
It's with the plus sign. (EDIT: See the 2 posts below from <arildno> and me).
 
Last edited:
  • #3
Yeah, that's what I calculated but my calculator and textbook says otherwise...
 
  • #4
It depends on which interval arcsine maps [-1,1] onto.

Conventionally, arcsine maps this interval onto [itex]-\frac{\pi}{2},\frac{\pi}{2}[/itex].

In this case, we have [tex]\frac{d}{dx}arcsine(x)=\frac{1}{\sqrt{1-x^{2}}}[/tex], and it follows that the anti-derivative will use the +sign
 
  • #5
You may wonder where the range of "arcsin" comes into play. Remember that

[tex] \frac{d \mbox{arcsin} x}{dx} = \pm \frac{1}{\sqrt{1-x^2}} [/tex], because

in the computation of this derivative you meet a point where

[tex] \sin x = p \Rightarrow \cos x = \pm \sqrt{1-p^2} [/tex].

To remove the sign ambiguity, you have to choose an appropriate interval either for the domain or for the range of the functions involved.
 
  • #6
Thanks for the replies arildno and bigubau.

@Bigubau, so do you mean this step when differentiating arcsin(x):

let [tex]y= arcsin(x) , sin(y)=x[/tex] and

differentiating [tex]sin(y)= x[/tex],

[tex]cos(y) \frac{dy}{dx} =1 [/tex]

=> [tex]\frac{dy}{dx}=\frac{1}{cos(y)}=*\frac{1}{\sqrt{1-sin^2(y)}}=*\frac{1}{\sqrt{1-x^2}}[/tex]

* where it should be [tex]cos(y)= \pm\sqrt{1-sin^2(y)}=\pm\sqrt{1-x^2}[/tex]

When I integrate arcsin(x) using integration by parts, I get

[tex] xarcsin(x) - \int \frac{x}{\sqrt{1-x^2}} dx [/tex]

so it comes out with the positive [tex]+\sqrt{1-x^2}[/tex]

Is there another method to integrate the last part? Apparently, when I calculate a definite integral, it's got to be either plus or minus and my calculator and textbook demands the minus. For example they both give

[tex]\int_{0}^{1} arcsin(x) = \frac{\pi}{2}-1[/tex]



Thanks for the support,
Charismaztex
 

1. What is the definition of arcsin(x)?

The arcsine function, denoted as arcsin(x) or sin-1(x), is the inverse function of the sine function. This means that for any value of x, the arcsine function will give the angle whose sine is x. In other words, arcsin(x) = y if and only if sin(y) = x.

2. What is the domain and range of arcsin(x)?

The domain of arcsin(x) is the set of real numbers between -1 and 1, inclusive. This is because the sine function only takes on values between -1 and 1, and the arcsine function is the inverse of the sine function. The range of arcsin(x) is the set of real numbers between -π/2 and π/2, inclusive.

3. How do you graph arcsin(x)?

The graph of arcsin(x) is a reflection of the graph of the sine function over the line y=x. This means that the x and y coordinates are switched. The graph of arcsin(x) is a curve that starts at (-π/2, -1) and ends at (π/2, 1), and it is symmetrical about the point (0, 0).

4. What are some real-world applications of arcsin(x)?

The arcsine function is commonly used in physics and engineering, particularly in the fields of mechanics, waves, and oscillations. For example, it can be used to calculate the angle of a projectile launched at a certain velocity, or the amplitude of a sound wave with a given frequency. It is also used in trigonometric identities and calculus problems.

5. How do you solve equations involving arcsin(x)?

To solve equations involving arcsin(x), you can use algebraic manipulation and inverse trigonometric identities. For example, if you have an equation sin(x) = 0.5, you can take the arcsine of both sides to get x = arcsin(0.5) = π/6 or 30 degrees. It is important to pay attention to the domain and range of the arcsine function when solving equations, as it will affect the possible solutions.

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