Evaluate $\int \sqrt{\frac{\cos x - \cos^3 x}{1-\cos^3 x}}dx$
My attempt:
$I=\int \sqrt{\frac{\cos x - \cos^3 x}{1-\cos^3 x}}dx=\int \sqrt{\frac{\cos x(1-\cos^2 x)}{1-\cos^3 x}}dx=\int \sqrt{\frac{\cos x\sin^2 x}{1-\cos^3 x}}dx=\int \sin x\sqrt{\frac{\cos x}{1-\cos^3 x}}dx=\int\sqrt{\frac{\cos x}{1-\cos^3 x}}(\sin x)dx=-\int \sqrt{\frac{\cos x}{1-\cos^3 x}}(-\sin x)dx$
Let $z=\cos x$
$\therefore dz=(-\sin x)dx$
$\therefore I=-\int\sqrt{\frac{z}{1-z^3}}dz$
I cannot understand how to proceed further. Please help.
Hint: Substitute $u=z^{3/2}$. You should then be able to recognise the integral you get.
How could you have come up with this yourself? Well note that you have $\sqrt z dz$ on top (which is the same up to a constant as $d(z^{3/2})$), and you have $1-z^3$ on the denominator, whereas you're probably more used to seeing things of the form $1-z^2$ on the denominator. So the substitution $z^{3/2}$ has the effect of transforming the denominator into something you recognise, while also getting rid of the squareroot on the numerator.