The integration of arcsin(x)

1. Dec 29, 2009

Charismaztex

Hello all,

I've solved this equation to get:

$$\int \arcsin{(x)}dx = \sqrt{1-x^2}+ x\arcsin{(x}) + C$$

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

$$-\sqrt{1-x^2}$$

$$\int \arcsin{(x)}dx = -\sqrt{1-x^2}+ x\arcsin{(x}) + C$$

This minus sign has been confirmed by a website I came across
Code (Text):
[PLAIN]http://math2.org/math/integrals/tableof.htm
[/PLAIN] [Broken]
Elsewhere on this forum I've seen the answer with the plus sign:
Code (Text):
giving rise to this
inconsistency.

Any help is appreciated and thanks in advance,
Charismaztex

Last edited by a moderator: May 4, 2017
2. Dec 29, 2009

dextercioby

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

Last edited: Dec 29, 2009
3. Dec 29, 2009

Charismaztex

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

4. Dec 29, 2009

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

5. Dec 29, 2009

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.

6. Dec 29, 2009

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