This sequence of functions looks simple but

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Homework Statement




The question is attached in the picture.

The Attempt at a Solution



Since V0 = 1,

Thus V1 = \stackrel{1}{2}∏R2 which is a constant.

Then shouldn't Vn be as in the picture?
 

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I think there is a typo in the question.

(\sqrt{R^2 - x^2}) should be (\sqrt{R^2 - x^2})^n
 
Sourabh N said:
I think there is a typo in the question.

(\sqrt{R^2 - x^2}) should be (\sqrt{R^2 - x^2})^n

hmm are u sure about that?
 
Yes.
 
unscientific said:

Homework Statement




The question is attached in the picture.

The Attempt at a Solution



Since V0 = 1,

Thus V1 = \stackrel{1}{2}∏R2 which is a constant.

Then shouldn't Vn be as in the picture?

V_1(R) = <br /> \int_{-R}^R V_0 \left(\sqrt{R^2-x^2} \right) \, dx = \int_{-R}^R \; 1 \, dx = 2R,
etc.

RGV
 
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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