# determine $\int x\sqrt{1-x^2}\,dx$

I have to determine $\int x\sqrt{1-x^2}\,dx$ and I have a little question about the substitution. I tried to subsitute $t=1-x^2$. It is $dt=-2xdx$ and therefore $dx=\frac{-dt}{2x}$. But it is the following calculation allowed: $\int x\sqrt{1-x^2}\,dx=\int x\sqrt{t}\,\frac{-dt}{2x}=\int \frac{-1}{2}\sqrt{t}\,dt$ ? (I'm not sure if it is ok to write $\int x\sqrt{t}\,\frac{-dt}{2x}$). Regards

• Yes, it is absolutely ok to write that, provided you remember the functional relation between $x$ and $t$. (Don't take $x$ or $t$ for a constant.) – Yves Daoust Apr 24 '15 at 13:17

You can do as the following :

Since $xdx=-\frac{1}{2}dt$, $$\int x\sqrt{1-x^2}dx=\int\left(\sqrt{1-x^2}\right)\cdot \color{red}{xdx}=\int \sqrt t\cdot\color{red}{\frac{-1}{2}dt}$$

An idea:

We have

$$\int\sqrt x\;dx=\frac23 x^{3/2}+C\implies \int f'(x)\sqrt{f(x)}dx=\frac23 (f(x))^{3/2}+C$$

Now, just observe that

$$\;x=-\frac12\left(1-x^2\right)'\;$$

For complete safety, you can write

$$\int x\sqrt{1-x^2}\,dx=\int x(t)\sqrt{t}\,\frac{-dt}{2\,x(t)}=-\frac12\int \sqrt{t}\,dt.$$

Had the $x$'s not canceled each other, you should have had to substitute $x$ by $\sqrt{1-t}$ in the end.