To solve the integral:

$\int\frac {x}{\sqrt{4-x^2}}\ dx$

I used the substitution:

$u = \sqrt{4-x^2}$


$\frac{du}{dx} = -\frac {x}{\sqrt{4-x^2}}$

$dx = -\frac {\sqrt{4-x^2}}{x}\ du$

So the integral becomes:

$\int\frac {x}{u}(-\frac {\sqrt{4-x^2}}{x})\ du$

$= \int\frac {x}{u}(-\frac {u}{x})\ du$

$= \int -1\ du$

$= -u+c$

$= -\sqrt{ 4-x^2}+c$

However the solution I have seen is:

Use the substitution $x = 2\sin(t)$, so $dx = 2\cos(t)\ dt$

So the integral becomes:

$\int \frac {2\sin(t)(2\cos(t))}{\sqrt{\cos(t)^2}}\ dt$

After some rearranging the integral becomes:

$\int 2\sin(t)\ dt$

Which gives:


Then using $\sin(t) = \frac x2$:

$\cos(t)^2 = 1 - \sin(t)^2$

$\cos(t)^2 = 1 - \frac {x^2}{4} = \frac{4-x^2}{4}$

$\cos(t) = \frac{\sqrt{4-x^2}}{2}$

Which then gives the solution as:

$= -\sqrt{ 4-x^2}+c$

I feel like the method I used was more efficient as it didn't require the use of trig functions. I'm fairly confident my method is correct so why would the solution use trig functions? Was I just lucky that I decided $u = \sqrt{ 4-x^2}$ was a good substitution?

  • 1
    $\begingroup$ substitution with trig function is more universal not only for this example but for any function $f(x,\sqrt{a^2-x^2})$ $\endgroup$
    – aid78
    Commented Jan 9, 2018 at 20:25

3 Answers 3


Well, it is no harm in solving the question either way. Some are more efficient, some are not. What matters is, it should be correct, which in this case, it is.

By the way, the trigonometric substitution is standard in these type of problems, (involving $\sqrt{a^2\pm x^2}$) so you should keep in mind to use it first, because it works most of the times.


Let $4-x^2=u$

Then $xdx=-\frac {du} 2$......


The two have more in common than they first seem to because $u=\sqrt{4-4\sin^2 t}=\cos t,\,du=-\frac{x}{2}dt$. A trigonometric substitution is so useful in such a variety of problems it's become a standard part of the toolkit.

  • $\begingroup$ JG is it x or t ? in $du=-\frac{t}{2}dt$ $\endgroup$ Commented Jan 9, 2018 at 21:31
  • $\begingroup$ @Isham Since $x$ was the original variable, I'm comparing the two substitutions. $\endgroup$
    – J.G.
    Commented Jan 9, 2018 at 22:14
  • $\begingroup$ Ok sorry for the edit then $\endgroup$ Commented Jan 9, 2018 at 22:16

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