How Accurate is the Calculation of This Curve's Length?

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1. On page 1, the integral for L is taken w.r.t. the variable t, so you are missing 'dt'.
2. On page 1, the first eqn. under the integral for L is not dx/dt. However, the expression below that appears to be OK.
3. On page 2 at the bottom, you have not integrated 8t correctly w.r.t. the variable 't'. If you integrals don't have a d(something) somewhere, you don't have a complete expression.
 
thanks so much

so the last step for L is correct except integration of 8t
 
it appears so.
 
Is the final answer is (pi^2 )
 
manal950 said:
Is the final answer is (pi^2 )
Yes.

As mentioned at least three other times by other people, your integrals need the differential, dt or dx or whatever it happens to be. If you persist in leaving them out, you will get the wrong answer when you attempt to evaluate integrals using techniques such as trig substitution and integration by parts.
 
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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