Derivative of Integrals: Calculating with the Fundamental Theorem of Calculus

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What is the result of this derivative: \frac{d}{da}\int^{\infty}_{a} f_{1}(ax)f_{2}(x)dx
 
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You can write, say, g(a, b) = \int_a^\infty f_1(bx)f_2(x) \,dx; then using the chain rule you get \frac{d}{da} g(a, a) = g_1(a, a) + g_2(a, a), where g1 is the partial derivative of g with respect to the first argument, and similarly for g2. For calculating g1, use the fundamental theorem of calculus; for calculating g2, move the derivative under the integral sign.
 
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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