Tricky Integral, U-substitution, or Trig Integral?

Question

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:

$$\int\frac{du}{\root4\of{u}\sqrt{1-u}}$$

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 $x=\sin\theta,\:dx=\cos\theta d\theta$

thus

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

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

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,

Thanks

Answer 1

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 :)

Answer 2

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...