What is the Correct Integral for x * (arcsin x) * (1-x2)-1/2 dx?

[Zoe-b]

Homework Statement

Find the integral of x * (arcsin x) * (1-x²)^-1/2 dx

Homework Equations

integration by parts

The Attempt at a Solution

u = x, u' = 1
v' = (arcsin x) * (1-x²)^-1/2 (= f(x) * f'(x) )
so v = ((arcsin x)²) / 2

using integration by parts
uv - integral of u'v

= x * ((arcsin x)²) / 2 - integral of ((arcsin x)²) / 2


Now use integration by parts for a second time to find the new integral, taking the half out as a constant:


w = arcsin x w' = (1-x²)^-1/2

z' = arcsin x
so z = x * (arcsin x) + (1-x²)^1/2

wz - integral of w'z

= x * (arcsin x)² + (arcsin x) * (1-x²)^1/2 - integral of [
x * (arcsin x) * (1-x²)-^1/2 + 1]


Substitute back into first equation (ie multiply above by -1/2)

integral of x * (arcsin x) * (1-x²)^-1/2 dx =

x * ((arcsin x)²) / 2 - x * ((arcsin x)²) / 2 - ((arcsin x) * (1-x²)^1/2) / 2 + 1/2 * [integral of x * (arcsin x) * (1-x²)^-1/2 dx] + 1/2 * integral of 1 dx


let [integral of x * (arcsin x) * (1-x²)^-1/2 dx ] = I


I = - ((arcsin x) * (1-x²)^1/2) / 2 + I/2 + x/2

I/2 = - ((arcsin x) * (1-x²)^1/2) / 2 + x/2

I = - ((arcsin x) * (1-x²)^1/2) + x


However when I check this by differentiation I end up with - x * (arcsin x) * (1-x²)^-1/2. Hence I think the correct answer is ((arcsin x) * (1-x²)^1/2) - x

Thanks

 

[Mentallic]

Ironically, the mistake was in your differentiation

 

[Zoe-b]

Lol. Finally found where I was going wrong- differentiated (1-x²)^1/2 aand forgot the minus sign. Thought I was going mad :P Thanks!