Surface area of sin(x) rotated about the x-axis

emc92
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got stuck on doing the substitution.. any suggestions?
 

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emc92 said:
got stuck on doing the substitution.. any suggestions?

For one thing, the derivative of 1 + cos2(x) is -2sin(x)cos(x), not -2sin(x).

That substitution doesn't seem to help anyway.

Try u = cos(x) instead.
 
wow, i did not see that.
thanks so much!
 

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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