Why Is the Definite Integral Zero for Symmetric Functions?

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Homework Statement



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The Attempt at a Solution



I have no idea how to start this integral. Can anybody give me a hint to start it off?
Thank you.
 
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Sketch a graph. Anything strike you about the relation between the part of the graph for x<0 vs x>0?
 
Well I noticed that x>0 is the reflection of x<0 except with negative values. Does this mean that the answer is 0? Also is it even possible to solve this integral?
 
No, you aren't going to get far trying to find an indefinite integral. But the definite integral is zero. In general, if f(-x)=(-f(x)) then the integral over a symmetric interval around zero is zero. You can formally show this by splitting the integral into two parts and working out their relation by doing the substitution u=(-x).
 

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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