Arc length (mostly a problem with integration)

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Homework Statement



Find the arc length oh the graph f(x)=cosx on the integral [0,\frac{\pi}{2}]

Homework Equations



\int^{b}_{a}\sqrt{1+{f'(x)}^{2}}dx

The Attempt at a Solution



Alright so I took the derivative of f(x) to get f'(x)=-sinx then I squared it to get sin^{2}x so I could plug it into the formula to get \int^{\frac{\pi}{2}}_{0}\sqrt{1+sin^{2}x} the problem is when I try to integrate...well i can't. I tried to use substitution but since i don't have cos^{2}x anywhere I had some issues.
 
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Anybody? Please this is driving me crazy!
 
You can't do it. You have to do a numerical approximation
 
I get 1.910098938245763 with wxmaxima
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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