Question

Doing integrals with Residues at Infinity specifically with Complex Analysis $\int_0^1 \sqrt{x}\sqrt{1-x}dx$ .
Apparently if I split the analytic function f(z)= into $\sqrt{z}$ and $\sqrt{1-z}$. I am able to see where on the Complex Plane it is defined.
But then somehow this problem uses information with solving it via the Facts of Residue's at Infinity. Yes, it is a Real Integral but I am to solve it using Complex Analysis and Branch Cuts. And lastly the fact with Residues at Infinity since it is undefined from 0 to 1 and either can be used with Doggie Bone Contour or Residue's at Infinity.
The why and how it can be solved using Residue's at Infinity isn't imminently clear to me.

Answer 1

Given that integral_0^1 squareroot x squareroot 1 - x dx compute the indefinite integral integral squareroot x squareroot 1 - x dx apply - substitution: u = squareroot x = integral 2u^2 squareroot 1- u^2 du Take the constant out: integral af(x) dx = a integral f(x)dx = 2 integral u^2 squareroot 1 - u^2 du put u = sinv = 2 integral sin 2v cos 2v dv [therefore cos 2x sin 2x = 1 - cos(4x)/8] = 2 integral 1 - cos(uv)/8 dv Take constant out integral f(x) dx = a integral f(x)dx = 2 times 1/6 integral 1 - cos(uv)dv = 2 times 1/8(integral(1dx) - integral cos(ux)dv) = 2 times 1/8(v - 1/4 sin(4 times 1)) substitute back v = arc sin(u), u = squareroot x = 2 times 1/8(arc sin(squareroot x) - 1/4 sin(4 arc sin(squareroot 1))) = 1/4(arc sin squareroot x) - 1/4 sin (4 arc sin(squareroot x)) + c compute the boundaries integral_0^1 squareroot x squareroot 1 - x dx pi/8