Integral problem, does b & a change sometime in the problem?

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Problem number 16
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For the record the y at the end is cubed. Thanks in advance.
 
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If you are referring to the limits of integration, then yes they change if you make the proper substitution that makes this integral very easy.
 
What do you mean by: "do b and a change sometime?" Are you talking about the limits of integration changing?
 
Feldoh said:
What do you mean by: "do b and a change sometime?" Are you talking about the limits of integration changing?

Yes the bounds or limits. b=2 and a=0 Why do they change to other numbers later on during the problem?
 
If you substitute, for example, u=1+y^3, then the limits will change from ya=0, yb=2 to ua=1 and ub=9. That's because you're integrating with respect to u instead of y.
 
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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