An integration problem which is supposed to be easy

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I am just lost at how to use the given information to find out the required integral. I've tried fundamental theorem of calculus, substitution, and integration by parts, but they do not seem to work for me.

Thank you!
 
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Sure. Use the fundamental theorem in the form of the Liebniz integral rule. Then look for something to integrate by parts.
 
Actually, I think it's even easier if you change the integration variable to u=x-t. THEN apply the fundamental theorem.
 
Dick said:
Actually, I think it's even easier if you change the integration variable to u=x-t. THEN apply the fundamental theorem.

Using u=x-t, du=-dt, and the integral becomes -(x-u)*f(u) du integrating from x to 0, but now I am not sure how it could be integrated to some useful form relating it to 1-cos(x) ?
 
Your u integral is equal to 1-cos(x). Now differentiate both sides.
 
Dick said:
Your u integral is equal to 1-cos(x). Now differentiate both sides.

I got it, thank you very much!
 
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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