## The integration of arcsin(x)

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:
http://math2.org/math/integrals/tableof.htm
Elsewhere on this forum I've seen the answer with the plus sign:
Code:
http://www.physicsforums.com/showthread.php?t=89216
giving rise to this
inconsistency.

Any help is appreciated and thanks in advance,
Charismaztex

 PhysOrg.com science news on PhysOrg.com >> Heat-related deaths in Manhattan projected to rise>> Dire outlook despite global warming 'pause': study>> Sea level influenced tropical climate during the last ice age
 Blog Entries: 9 Recognitions: Homework Help Science Advisor It's with the plus sign. (EDIT: See the 2 posts below from and me).
 Yeah, that's what I calculated but my calculator and textbook says otherwise...

Recognitions:
Gold Member
Homework Help

## The integration of arcsin(x)

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

 Blog Entries: 9 Recognitions: Homework Help Science Advisor 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.
 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