How to integrate this function using arccos?

Question

$$-\int \frac{\mathrm dx}{\sqrt{x^3-x^2}}$$

I know I have to use the formulas for things like arccos and arccsin, but I don't know how to get it in that form.

Answer 1

I assume $x > 1$. Then your integral has the form

$$- \int \frac{\mathrm dx}{x\sqrt{x - 1}}.$$

Now let $u = \sqrt{x - 1}, \ \mathrm dx = 2 \sqrt{x - 1}\ \mathrm du$. This substitution leads to

$$- 2 \int \frac{\mathrm du}{u^2 + 1}.$$

But this is a standard integral, which evaluates to

$$- 2 \arctan(u) + C.$$

Undoing the substitution gives

$$\fbox{$-2\arctan(\sqrt{x - 1}) + C$.}$$

Answer 2

The integrand is $-1/(x\sqrt(1-x))$. Let $u^2=x$. $2udu=dx$.

=> integral $-2/(u\sqrt(u^2-1)) du = -2\text{arcsec}(u)+ C$

So our answer is $-2\text{arcsec}(\sqrt(x))+C$.

Answer 3

Let $x = \frac1{t^2}$. Then,

$$-\int \frac{\mathrm dx}{\sqrt{x^3-x^2}}=\int \frac {2dt}{\sqrt{1-t^2}}=-2\text{arccos} (t)+C$$

Answer 4

Start with $x=\sec^2 u$

$$\int\frac{dx}{\sqrt{x^3-x^2}}=\int\frac{dx}{x\sqrt{x-1}}=2\int \ du=2u=2\sec^{-1}(\sqrt{x})+C$$