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Zoe-b
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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