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

[Charismaztex]

Hello all,

I've solved this equation to get:

<br /> \int \arcsin{(x)}dx = \sqrt{1-x^2}+ x\arcsin{(x}) + C<br />

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

-\sqrt{1-x^2}

instead of a +, leaving the answer

<br /> \int \arcsin{(x)}dx = -\sqrt{1-x^2}+ x\arcsin{(x}) + C<br />

This minus sign has been confirmed by a website I came across

[PLAIN]http://math2.org/math/integrals/tableof.htm

[/PLAIN]
Elsewhere on this forum I've seen the answer with the plus sign:

https://www.physicsforums.com/showthread.php?t=89216

giving rise to this
inconsistency.

Any help is appreciated and thanks in advance,
Charismaztex

 

[dextercioby]

It's with the plus sign. (EDIT: See the 2 posts below from <arildno> and me).

 

[Charismaztex]

Yeah, that's what I calculated but my calculator and textbook says otherwise...

 

[arildno]

It depends on which interval arcsine maps [-1,1] onto.

Conventionally, arcsine maps this interval onto -\frac{\pi}{2},\frac{\pi}{2}.

In this case, we have \frac{d}{dx}arcsine(x)=\frac{1}{\sqrt{1-x^{2}}}, and it follows that the anti-derivative will use the +sign

 

[dextercioby]

You may wonder where the range of "arcsin" comes into play. Remember that

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

in the computation of this derivative you meet a point where

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

To remove the sign ambiguity, you have to choose an appropriate interval either for the domain or for the range of the functions involved.

 

[Charismaztex]

Thanks for the replies arildno and bigubau.

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

let y= arcsin(x) , sin(y)=x and

differentiating sin(y)= x,

cos(y) \frac{dy}{dx} =1

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

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

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

xarcsin(x) - \int \frac{x}{\sqrt{1-x^2}} dx

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

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

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



Thanks for the support,
Charismaztex