1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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.) $\endgroup$ – Yves Daoust Apr 24 '15 at 13:17
2
$\begingroup$

You can do as the following :

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

$\endgroup$
0
$\begingroup$

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

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

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.