2
$\begingroup$

I have to calculate $$\int \frac{dx}{\sqrt{(x+2)(3-x)}}$$.

I tried to use -

$\int u'v = uv - \int v'u$, but im pretty stuck.

Thanks.

$\endgroup$
2

5 Answers 5

4
$\begingroup$

Hint: rewrite the integral as $$ \int\frac{\mathrm{d}x}{\sqrt{\frac{25}4-\left(x-\frac12\right)^2}} $$ then substitute $\frac52\sin(u)=x-\frac12$.

$\endgroup$
2
$\begingroup$

$$\int\frac{1}{\sqrt{(x+2)(3-x)}}\space\text{d}x=\int\frac{1}{\sqrt{-x^2+x+6}}\space\text{d}x=\int\frac{1}{\sqrt{\frac{25}{4}-\left(x-\frac{1}{2}\right)^2}}\space\text{d}x=$$


Substitute $u=x-\frac{1}{2}$ and $\text{d}u=\text{d}x$:


$$\int\frac{1}{\sqrt{\frac{25}{4}-u^2}}\space\text{d}u=\int\frac{2}{5\sqrt{1-\frac{4u^2}{25}}}\space\text{d}u=\frac{2}{5}\int\frac{1}{\sqrt{1-\frac{4u^2}{25}}}\space\text{d}u=$$


Substitute $s=\frac{2u}{5}$ and $\text{d}s=\frac{2}{5}\space\text{d}u$:


$$\int\frac{1}{\sqrt{1-s^2}}\space\text{d}s=\arcsin\left(s\right)+\text{C}=\arcsin\left(\frac{2u}{5}\right)+\text{C}=$$ $$\arcsin\left(\frac{2\left(x-\frac{1}{2}\right)}{5}\right)+\text{C}=\arcsin\left(\frac{2x-1}{5}\right)+\text{C}$$

$\endgroup$
0
$\begingroup$

HINT:

We need $(x-3)(x+2)<0\iff-2<x<3$

Set $x=-2\cos^2y+3\sin^2y\iff x-3=-5\cos^2y,x+2=5\sin^2y$

$dx=?$

$\endgroup$
0
$\begingroup$

\begin{align} \int \frac{dx}{\sqrt{(x+2)(3-x)}} =& \ \text{sgn} (\frac12-x)\int \frac{d\left(\sqrt{(x+2)(3-x)}\right)}{\sqrt{\frac{25}4- (x+2)(3-x)}}\\ =& \ \text{sgn} (\frac12-x) \sin^{-1}\bigg(\frac25 \sqrt{(x+2)(3-x)}\bigg)+C \end{align}

$\endgroup$
2
  • $\begingroup$ Interesting that this should be equivalent to other arcsine representations $\endgroup$
    – FShrike
    Aug 11, 2022 at 21:07
  • $\begingroup$ @FShrike - indeed equivalent, less known though $\endgroup$
    – Quanto
    Aug 13, 2022 at 19:41
-1
$\begingroup$

You can $u$-substitute $u=x-1/2$ to get

$$\int \frac{du}{\sqrt{ (u+5/2)(5/2-u)}} = \int \frac{du}{\sqrt{25/4 - u^2}} $$

and trig substitutions take it from there. You get the idea to let $u = x-1/2$ because the average of $2$ and $-3$ is $1/2$ and it makes the factors more symmetric.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .