Expert: Paul Klarreich - 5/28/2007

Question
Function f(x) = sinx

I’m looking at the area underneath a sine graph with the reference points from the x axis of 0 to pi/4. So the area underneath the curve between those to points from the x axis.

So that would read 0 to pi/4 to the integral of sinx dx

Integral of sinx is –cosx

Therefore (-cos(pi/4)) – (-cos(0)) = Area



I am also looking at the area of the sine graph between curve to the y axis.
So my two reference points would be off the y axis.

0 to sqrt(2)/2

So that reads 0 to sqrt(2)/2 to integral of arcsiny dy

Accept I don’t know the antidifferentation of inverse sine
My question is how would one find the antiderivative of inverse sine. What would be the method or rule used to find the inverse sine.


What I know:

d/dx(arcsin) = 1/sqrt(1-x^2)
integral of 1/sqrt(1-x^2) dx = arcsin + C

What I don’t know:
Integral of arcsin = ?

Cheers James

Answer
Questioner:   James
Category:  Advanced Math

 
Subject:  Antidifferentation – area underneath a curve
Question:  Function f(x) = sinx

I’m looking at the area underneath a sine graph with the reference points from the x axis of 0 to pi/4. So the area underneath the curve between those two points from the x axis.

So that would read 0 to pi/4 to the integral of sinx dx

Integral of sinx is –cosx

Therefore (-cos(pi/4)) – (-cos(0)) = Area


I am also looking at the area of the sine graph between curve to the y axis.
So my two reference points would be off the y axis.

0 to sqrt(2)/2

So that reads 0 to sqrt(2)/2 to integral of arcsiny dy

Accept I don’t know the antidifferentation of inverse sine
My question is how would one find the antiderivative of inverse sine. What would be the method or rule used to find the inverse sine.

What I know:

d/dx(arcsin) = 1/sqrt(1-x^2)
integral of 1/sqrt(1-x^2) dx = arcsin + C

What I don’t know:
Integral of arcsin = ?

Cheers James
..............................................
Hi, James,

Some of the stuff you wrote is not detailed enough for me to comment on it, but it seems you just want to know how to do this integral:

{
| arcsin(x) dx
}

I don't think I can tell you.  That is information you are not allowed to have until you reach Calc II.  However, if you promise to destroy this message if you are not authorized to receive it, here it is:

. . . . . . .CLASSIFIED MESSAGE. . . . . . . . .

Use integration by parts:  [Never heard of it?  Then destroy this message.]
                        dx
u = arcsin(x),  du = -----------
                    sqrt(1-x^2)

dv = dx,   v = x

Now write:
             {     x dx
x arcsin(x) - | --------------  
             } sqrt(1 - x^2)

The second integral can now be handled by a basic substitution:

Let   u = 1 - x^2,  then  du = - 2x dx,  x dx = - du/2

             {     du
x arcsin(x) + | --------------  
             } 2 sqrt(u)

              1 {
x arcsin(x) + ---| u^-1/2  du
              2 }

              1  u^1/2
x arcsin(x) + --- -----
              2  1/2
The 1/2's cancel, and you have:

x arcsin(x) + sqrt(1 - x^2)

which you could look up in any standard table of integrals.  And which you can certainly check by differentiating.

You can also check THE INTEGRATOR.  That is not an action movie, but a web site:

http://integrals.wolfram.com/index.jsp

which allows you to enter integrals and works them out for you.