Solve Integral Inequality with cos(x): \int\sqrt{xcos(x)} dx 0..Pi/2

phyguy321
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Show that \int\sqrt{xcos(x)} dx from 0..Pi/2 \leq Pi/2 \sqrt{2}


Having problems with the cos(x) part. Maple gives -sqrt(2)*EllipticK((1/2)*sqrt(2))+2*sqrt(2)*EllipticE((1/2)*sqrt(2)) for the integral of the cos part.

what are EllipticK and EllipticE and how are they evaluated?
so lost right now
 
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cosx\leq 1=>xcosx<x=>\sqrt{xcosx}\leq \sqrt{x} now

\int_0^{\frac{\pi}{2}}\sqrt{xcosx}\leq \int_0^{\frac{\pi}{2}}\sqrt{x}=\frac{1}{2}\frac{2}{3}x^{\frac{3}{2}}|_0^{\frac{\pi}{2}}=\frac{1}{3}\frac{\pi}{2\sqrt{2}}<\frac{\pi}{2\sqrt{2}}
 
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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