given a volume of rotation problem, the result is the evaluation of the following integral

$$\int_{0}^{1/2} \frac{dx}{\root4\of{1-x^2}}$$

For our purposes here, let's just consider the indefinite integral

$$\int \frac{dx}{\root4\of{1-x^2}}$$

I've tried 2 techniques, the first is u substitution:

if we let $u=1-x^2,\:du=-2xdx,\: dx=-\frac{du}{2x},\: x=\sqrt{1-u},\:dx=-\frac{du}{2\sqrt{1-u}}$

Then we get the integral:


which seems to me to be a tougher integral to evaluate.

Next method to try, trig substitution

$$\int \frac{dx}{\root4\of{1-x^2}}$$is equal to $$\int \frac{dx}{\root4\of{1^2-x^2}}$$

which suggests the substitution $x=\sin\theta,\:dx=\cos\theta d\theta$


$$\int \frac{\cos\theta\:d\theta}{\root4\of{1-\sin^2\theta}}=\int \frac{\cos\theta\: d\theta}{\root4\of{(\cos\theta)^2}}$$

$$=\int \frac{\cos\theta\:d\theta}{\sqrt{\cos\theta}}=\int \sqrt{\cos\theta }\,d\theta$$

This also seems like a tough integral to evaluate...

This question is asked in my textbook before the section on trig integrals, so I am assuming there is a quick dirty trick used to evaluate an integral like this

Please share some wisdom,


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    $\begingroup$ Alpha gets an answer in terms of a hypergeometric function for the indefinite integral and reports a numeric rather than symbolic result for the definite integral. I wouldn't have much hope in solving it by hand. $\endgroup$ – Ross Millikan Feb 23 '16 at 21:42
  • $\begingroup$ seems odd that this question would be asked in a first year calculus textbook.. in a chapter before the numerical integration section $\endgroup$ – helpmeh Feb 23 '16 at 21:44
  • $\begingroup$ What was the original question? The one from which this integral arose from? $\endgroup$ – Patrick Da Silva Feb 23 '16 at 21:47
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    $\begingroup$ In exact words.. Let R be the region bounded by the following curves, use the disk method to find the volume of the solid generated when R is revolved about the x axis #23 $$y=\frac{1}{\sqrt[4]{1-x^2}}. \: y=0,\: x=0,\:x=1/2$$ $\endgroup$ – helpmeh Feb 23 '16 at 21:47

The disk method around the $x$-axis does not ask that you integrate $y$ but rather $\pi y^2$, because $y$ is serving as the radius of your disk.

This integral is much easier, as the work in your question shows :)


Evaluating the indefinite integral leads to a hypergeometric function ($x \, _2F_1\left(\frac{1}{4},\frac{1}{2};\frac{3}{2};x^2\right)$), and evaluating this at $1/2$ gets a special value of the hypergeometric function, so I don't think there is anything nice, so you might have got the rest of the question wrong to get to this...

  • $\begingroup$ after double and triple checking, the question in the textbook is exactly as i've posted it. I'll bring this to my prof's attention then, thanks $\endgroup$ – helpmeh Feb 23 '16 at 21:47

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