How Do You Solve This Challenging Integral Involving Trigonometric Substitution?

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Homework Statement


integrate the following function: x*√(2x-x2)


Homework Equations


substitutions?


The Attempt at a Solution


I substituted x with x=2sin2u. From that I ended up with ∫x*√(2x-x2)= 16∫sin4u*cos2u

Now I'm supposed to use a formula from an integrals table.. but which?
 
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You should complete the square under the sqrt. \displaystyle{\sqrt{1-(1-x)^2}} and then make the natural substitution.
 
edit: nvm
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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