Integrating $\int_0^1 \frac{1}{\sqrt{1-y^4}}dy$

Question

What do I need to manipulate to show that $$\int_0^1 \frac{1}{\sqrt{1-y^4}}dy=\frac 1 4 \int_0^1 t^{-3/4}(1-t)^{-1/2}dt$$

you are missing some constant... Just make a substitution $t=y^4$ — leshik, Aug 28 '13 at 17:17

Cameron Buie · Accepted Answer · 2013-08-28T17:23:44.373

5

Put $t=y^4.$ What then is $dt$ in terms of $y$ and $dy$? Solve that equation for $dy$ and substitute $y=t^{1/4}$ to finish.

Why did I make that choice? Well, $$\frac1{\sqrt{1-y^4}}=(1-y^4)^{-1/2},$$ so it was a fair guess that that substitution might work, and it did.

edited Aug 28 '13 at 17:23

answered Aug 28 '13 at 17:18

Cameron Buie

Glad we don't have to compute that integral.... – imranfat Aug 28 '13 at 18:16
That integral's just $$\frac14B\left(\frac14;,;\frac12\right);,;B=\text{the Beta function}$$ – DonAntonio Aug 28 '13 at 18:23

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