Expert: Steve Holleran - 2/27/2007

Question
I am trying to find the integral of xarcsinx dx  I know you would do it by parts but am stuck.  Could you help me out?

Answer
Hello Peter,

Okay, first, I'm sorry it took me awhile to get back to you, but its been awhile since I've used these methods, and I  finally got it to come out, but I'll tell you up front that I've got a sign mistake somewhere that I can't find.  Maybe you will be able to pinpoint it.

Anyway, you are correct that parts is the method to use, but after that initial step, there's an integral that has to be done by trig substitution.

To get started, if you take u=arctanx and dv = x dx, then
du = 1/sqrt(1-x^2) dx and v = x^2 / 2 , then integrating by parts gives:

int(x*arctanx dx) =

     x^2 / 2 * arcsin x - int[x^2 /2 * 1/ sqrt(1-x^2)*dx]

or  1/2 [ x^2 * arcsin x - int[(x^2 * dx)/sqrt(1-x^2)]


Okay, now just focusing on the second integral, this is where you need to make a trig substitution:

Take x = Sin A   so dx = Cos A dA.

Also, if you diagram a rt triangle with acute angle A, label the opposite side x and the hypotenuse 1, then the remaining leg is sqrt(1-x^2), and this is Cos A, also.


Then the integral becomes :

int[(Sin^2 A * cos A * dA) / cos A] and the cos A cancels out.  so you're left with:

int[ Sin^2 a * dA]   You have to use the trig identity

Sin^2 A = 1/2 *(1-cos 2A) now to replace Sin^2 A:

int[ 1/2 * (1-cos 2A) * dA]  separates into two integrals,

int[1/2 * dA] - int [1/2* cos 2A * dA] and you get

1/2 * A - 1/4 * sin 2A  .   

Now use the double angle formula for the sin 2A :

1/2 * A - 1/4 * 2 * sin A * cos A  and substitute back using the original trig equations

1/2 * arcsin x - 1/2 * x * sqrt(1-x^2).

Putting it all together with the first part of the result from integrating by parts (I hope you're still with me!) you should have the following as the result:


1/2 * x^2 * arcsin x - 1/4 * arcsin x - 1/4 * x * sqrt(1-x^2)

If you factor out 1/4, you can write it as :

1/4 [ 2x^2 * arcsin x - arcsin x - x * sqrt(1-x^2) or

1/4 * [(2x^2-1)arcsin x - x * sqrt(1-x^2).

Whew!!!

Now the only problem is this:  I used an online integrator to check this -- it uses Mathematica, and did it in about 0.4 seconds-- and it shows the sign in front of the
x * sqrt(1-x^2) term in the parenthesis as a + instead of a -.

I spent over an hour checking, and I couldn't find where I went wrong.  So, I apologize for this, but I thought the rest might be helpful to you.  If you figure it out, feel free to come back and let me know.

I hope this helps you out, Peter.

Steve Holleran